Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm. -Maths 9th

Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm. -Maths 9th
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Using the formulas A=s(s﹣a)(s﹣b)(s﹣c) s=a+b+c 2Solving forA A=1 4﹣a4+2(ab)2+2(ac)2﹣b4+2(bc)2﹣c4=1 4·﹣124+2·(12·6)2+2·(12·15)2﹣64+2·(6·15)2﹣154≈34.19704cm²
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Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Description : Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Last Answer : We have, Three sides13cm,13cm and 20cm. By using Heron's formula We need to get the semi-perimeter s= 2 a+b+c​ = 2 13+13+20​ = 2 46​ =23 Now, put the heron's formula, s= s(s−a)(s−b)(s−c)​ = 23(23−13)(23−13)(23−20)​ = 23×10×10×3​ =10 23×3​ =83.07cm 2

Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

Description : Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

Last Answer : Step-by-step explanation: ◾As we have given the two sides of triangle, let the three sides of triangle are (a) , (b), (c) . ◾And perimeter of given triangle is 10.5 cm ◾were, let us assume the sides are, ... . ◾So, the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 [Area ]=

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. Draw a line segment BC of length 4.8 cm. From B, point A is at a distance of 3.6 cm. ... at P. Joining BP, we obtain angle bisector of ∠B. Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = ½ x 139° = 19.5°

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°

The sides of a triangle are 8 cm, 15 cm and 17 cm. Find its area. -Maths 9th

Description : The sides of a triangle are 8 cm, 15 cm and 17 cm. Find its area. -Maths 9th

Last Answer : Let a = 8cm, b = 15cm, c = 17cm s = (a + b + c)/2 = (8 + 15/ + 17)/2 = 40/2 = 20cm ∴ Area = root under √s(s - a)(s - b)(s - c) = root under √20(20 - 8)(20 - 15)(20 - 17) = root under √20 x 12 x 5 x 3 = 60 cm2

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Description : Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Last Answer : Let the sides of the triangle be a = 12x cm, b = 17x cm, c = 25x cm Perimeter of the triangle = 540 cm Now, 12x + 17x + 25x = 540 ⇒ 54x = 54 ⇒ x = 10 ∴ a = (12 x10)cm = 120cm, b = (17 x 10) cm = 170 cm and c = (25 x 10)cm = 250 cm Now, semi-perimeter, s = 5402cm = 270 cm

One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th

Description : One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th

Last Answer : Perimeter of the largest (outermost) equilateral triangle = 3 24 = 72 cm. Now, the perimeter of the triangle formed by joining the midpoints of a given triangle will be half the perimeter of the original triangle. ∴ Required sum = 72 + ... -rac{1}{2}}\) = \(rac{72}{rac{1}{2}}\) = 72 x 2 = 144 cm.

If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Description : If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Last Answer : Let one of the remaining sides be x cm.Then, other side = \(\sqrt{5^2-x^2}\) cm∴ Area = \(rac{1}{2} imes{x} imes\sqrt{25-x^2}\) = 6⇒ \(x\sqrt{25-x^2}\) = 12 ⇒ x2(25 - x2) = 144⇒ 25x2 - x4 = 144 ⇒ x4 - 25x2 ... (x2 - 16) (x2 - 9) = 0 ⇒ x2 = 16 or x2 = 9 ⇒ x = 4 or 3∴ The two sides are 4 cm and 3 cm.

The lengths of the sides of a triangle are 7 cm, 13 cm and 12 cm. -Maths 9th

Description : The lengths of the sides of a triangle are 7 cm, 13 cm and 12 cm. -Maths 9th

Last Answer : Let, a = 7 cm, b = 13 cm, c = 12 cm ∴ s = (a + b + c)/2 = (7 +13 +12)/2 = 32/2 = 16 cm Area of △ ABC = under root( √s(s -a) (s - b)(s -c)) = under root( √16(16 - 7)(16 - 13)(16 - 12) = ... 24 √3 cm2 Also, Area of △ ABC = 1/2AC.BD 24 √3 = 1/2 x 12 x BD ⇒ BD = (24 √3 x 2)/12 = 4 √3 cm

The lengths of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area (in sq. cm) of this triangle is -Maths 9th

Description : The lengths of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area (in sq. cm) of this triangle is -Maths 9th

Last Answer : (b) 72 cm2Here sm = \(rac{9+12+15}{2}\) = 18 cm, where lengths of medians are m1 = 9 cm, m2 = 12 cm, m3 = 15 cm.∴ Area of triangle = \(rac{4}{3}\sqrt{18(18-9)(18-12)(18-15)}\) cm2= \(rac{4}{3}\sqrt{18 imes9 imes6 imes3}\) cm2 = \(rac{4}{3}\) x 9 x 6 cm2 = 72 cm2.

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Last Answer : For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ ... be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

Description : The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

Last Answer : The area of the triangle is

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Description : The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

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The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

Description : The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

Last Answer : The area of the triangle is

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Description : The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Last Answer : s= 2 4+4+2​ =5 Area of the triangle Δ= s(s−a)(s−b)(s−c)​ = 5(5−4)(5−4)(5−2)​ = 15​ cm 2

Find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm. -Maths 9th

Description : Find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm. -Maths 9th

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In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45º. The area of the triangle is: -Maths 9th

Description : In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45º. The area of the triangle is: -Maths 9th

Last Answer : (c) 25√2 cm2.ΔABC is an isosceles triangle with AB = AC = 10 cm. ∠A = 45° ∴ Area of ΔABC= \(rac{1}{2}\) x 10 x 10 x sin 45°[Using Δ = \(rac{1}{2}\) bc sin A]= \(rac{50}{\sqrt2}\) = \(rac{50}{\sqrt2}\) x \(rac{\sqrt2}{\sqrt2}\) = 25√2 cm2.

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Description : A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

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A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Description : A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Last Answer : Since, the given right angled triangle is revolved about the side 8 cm, it will form a Cone of radius 6cm and height 8cm. Volume of a cone = 1/3∏r2h = 1/3 3.14 6 6 8 = 301.44 cm3 Curved Surface area of a cone ... value of l in (i), we get Curved Surface area of a cone = 3.14 6 10 = 188.4 cm2

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Description : A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Last Answer : Since, the given right angled triangle is revolved about the side 8 cm, it will form a Cone of radius 6cm and height 8cm. Volume of a cone = 1/3∏r2h = 1/3 3.14 6 6 8 = 301.44 cm3 Curved Surface area of a cone ... value of l in (i), we get Curved Surface area of a cone = 3.14 6 10 = 188.4 cm2

If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Description : If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Last Answer : (b) 7.5 cm.Area of a triangle = \(rac{1}{2}\)x base x height= In-radius x semi-perimeter of the Δ \(\big[ ext{Using r =}rac{\Delta}{s}\big]\)Let the sides of triangle be 4x, 5x and 6x respectively. Given: In-radius = 3 ... x \(rac{15x}{2}\) = \(rac{6x}{2}\) x h ⇒ h = \(rac{45}{6}\) = 7.5 cm.

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Description : Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Last Answer : Steps of Construction (i) Draw BC = 12 cm. (ii) Construct ÐCBY = 90°. (iii) From ray BY, cut-off line segment BD = 18 cm. (iv) Join CD. (v) Draw the perpendicular bisector of CD intersecting BD at A. (vi ... = AC Now, BD = BA + AD ⇒ BD = AB + AC Hence, △ABC is the required triangle.

Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th

Description : Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th

Last Answer : Area of a trapezium = 1/2(sum of parallel sides) x (perpendicular diatance between them) = 1/2(25 + 13) x 8 = 152cm2

What is the volume of a right prism standing on a triangular base of sides 5 cm, 5 cm and 8 cm whose lateral surface area is 828 cm^2 ? -Maths 9th

Description : What is the volume of a right prism standing on a triangular base of sides 5 cm, 5 cm and 8 cm whose lateral surface area is 828 cm^2 ? -Maths 9th

Last Answer : Lateral surface area of a prism = Perimeter of base Height ⇒ 840 = (5 + 5 + 8) Height ⇒ Height = 8401884018 = 46 cm. = Semi perimeter of the triangular base = 182182 = 9 cm ∴ Area of triangle = 9(9- ... 4 1 = 12 cm2 ∴ Required volume of prism = Area of base Height = (12 46) cm3 = 552cm3

If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

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If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Description : A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Last Answer : NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula Ex 12.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths ... = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m S

The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Description : The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Last Answer : The other two sides of the triangle are 12 cm and 5 cm Explanation: Let the other two sides of triangle be x and y It's hypotenuse is 13 cm Perimeter of triangle = Sum of all sides ... When y = 12 x=17-y = 17-12 =5 So, the other two sides of the triangle are 12 cm and 5 cm

If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Description : If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Last Answer : No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 ... , the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Description : If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Last Answer : No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 ... , the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Description : A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Last Answer : Let given right triangle be ABC. Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps 1.Draw the base BC = 3.5 cm 2.Make ... AB = BD - AD = BD - AC [from Eq. (i)] => BD = AB + AC Thus, our construction is justified.

The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Description : The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Last Answer : s= 2 a+b+c​ = 2 35+54+61​ =75 Area, A= s(s−a)(s−b)(s−c)​ = 75(75−35)(75−54)(75−61)​ =420 5​ cm 2 Now, Area of the triangle is also given as A= 2 1​ ×a×h Where, h is the longest altitude. Therefore, 2 1​ ×a×h=420 5​ Hence, h=24 5​ cm

A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Description : A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Last Answer : Let given right triangle be ABC. Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps 1.Draw the base BC = 3.5 cm 2.Make ... AB = BD - AD = BD - AC [from Eq. (i)] => BD = AB + AC Thus, our construction is justified.

The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Description : The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th

Last Answer : The length of its longest altitude

Is it possible to construct a triangle with lengths of its sides as 7 cm, 8 cm and 5 cm? Give reason for your answer. -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 7 cm, 8 cm and 5 cm? Give reason for your answer. -Maths 9th

Last Answer : Solution :- Yes, because in each case sum of two sides is greater than the third side.

A right triangle ABC with sides 5 cm, -Maths 9th

Description : A right triangle ABC with sides 5 cm, -Maths 9th

Last Answer : Let ABC be a right triangle with AB = 12 cm, BC = 5 cm and AC = 13 cm. When △ABC is revolved about AB, it forms a right circular cone of radius BC = 5 cm and height AB = 12 cm. Thus, volume of cone formed = 1/3 πr2h = 1/3 x π x 52 x 12 = 100π cm3

If the sides of a triangle are 3 cm, 4 cm and 5 cm, then what is the radius of the circum-circle? -Maths 9th

Description : If the sides of a triangle are 3 cm, 4 cm and 5 cm, then what is the radius of the circum-circle? -Maths 9th

Last Answer : Semi-perimeter of triangle (s) = \(rac{3+4+5}{2}\)cm = 6 cm∴ Area of triangle A = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{6 imes3 imes2 imes1}\) cm2 = 6 cm2∴ Radius of circum-circle = \(rac{abc}{4( ext{Area of}\,\Delta)}\) = \(rac{3+4+5}{4 imes60}\) cm = 2.5 cm

The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Last Answer : area of parallelogram=2× area of triangle ABC =2×15=30sq cm theorem on area

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB=6cm, BC=8cm, CD=12cm and DA=14 cm. Now. Join AC. We have, ABC is a right angled triangle at B. Now, AC2=AB2+BC2 [by Pythagoras theorem]Now, AC2=AB2+BC2 ... =24(1+6-√)cm2=24+246=24(1+6)cm2 Hence, the area of quadrilateral is 241+6-√−−−−−−√cm2241+6cm2 .

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm. Now, join AC.

ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]

Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Last Answer : We know that, each angle of a rectangle is right angle (i.e., 90°) and its opposite sides are equal and parallel. To construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm, use the ... AD and CD. Thus, ABCD is the required rectangle with adjacent sides of length 5 cm and 3.5 cm.

Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°

Sides of a triangle are in the ratio of 12:17:25 and... -Maths 9th

Description : Sides of a triangle are in the ratio of 12:17:25 and... -Maths 9th

Last Answer : Let the sides of the triangle be 12x, 17x and 25x Perimeter of the triangle = 540 cm ∴ 12x + 17x + 25x = 540 ⇒ 54x = 540 ⇒ x = 10 Let a = 12x = 12 x 10 = 120 cm b = 17x = 17 x 10 = 170 cm ... ( √270 x 150 x 100 x 20) = 100 under root( √27 x 15 x 20) = 100 x 9 x 5 x 2 = 9000 cm2

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : NEED ANSWER

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : This answer was deleted by our moderators...

Find the area of a triangle whose vertices are (1, 3), (2, 4) and (5, 6). -Maths 9th

Description : Find the area of a triangle whose vertices are (1, 3), (2, 4) and (5, 6). -Maths 9th

Last Answer : Let OR = \(x\) units (i) ΔQOR ~ ΔPAR⇒ \(rac{PA}{AR}\) = \(rac{QO}{OR}\) ⇒ \(rac{6}{x+4}\) = \(rac{3}{x}\)⇒ \(rac{x}{x+4}\) = \(rac{3}{6}\) ⇒ \(rac{x}{x+4}\) = \(rac{1}{2}\)⇒ 2\(x\) = \(x\) + 4 ⇒ \(x ... = \(rac{1}{2}\) (OQ + AP) x OA = \(rac{1}{2}\) (3+6) x 4 = \(rac{1}{2}\) x 9 x 4 = 18 sq. units.

If the distance from the vertex to the centroid of an equilateral triangle is 6 cm, then what is the area of the triangle? -Maths 9th

Description : If the distance from the vertex to the centroid of an equilateral triangle is 6 cm, then what is the area of the triangle? -Maths 9th

Last Answer : (b) 27√3 cm2.Let G be the centroid of ΔPQR. Then, PG = 6 cm.Now, \(rac{PG}{GS}\) = \(rac{2}{1}\) ⇒ GS = 3 cm∴PS = PG + GS = 9 cm (i)∴ If a is the length of a side of ΔPQR, then ... = 6√3 cm∴ Area of equilateral ΔPQR = \(rac{\sqrt3}{4}\) (a)2= \(rac{\sqrt3}{4}\) x (6√3)2 cm2 = 27√3 cm2.