If two sides of one triangle are equal to two sides of another triangle and the contained angles are supplementary, show that the two sides are equal in area -Maths 9th

If two sides of one triangle are equal to two sides of another triangle and the contained angles are supplementary, show that the two sides are equal in area -Maths 9th
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If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.
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Description : Prove that angles opposite to equal sides of a triangle are equal. -Maths 9th

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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]

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

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The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

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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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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

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

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, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

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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

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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.

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Description : Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

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Last Answer : each interior opposite angles are 55

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An exterior angle of a triangle is 110° and the two interior opposite angles are equal find the interior opposite angels -Maths 9th

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Description : If one angle of a triangle is equal to the sum of the other two angles, then the triangle is -Maths 9th

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An exterior angle of a triangle is 110° and its two interior opposite angles are equal. Find each of these equal angles. -Maths 9th

Description : An exterior angle of a triangle is 110° and its two interior opposite angles are equal. Find each of these equal angles. -Maths 9th

Last Answer : 2x=110 X=55 x+x+y=180 110+y=180 Y=70

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Let A, B, C be the angles of a plain triangle (A/2) = (1/3 ), tan (B/2) = (2/3). Then tan (C/2) is equal to -Maths 9th

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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.

State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

In the diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. -Maths 9th

Description : In the diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. -Maths 9th

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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

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

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 ]=

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

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

Last Answer : 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

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 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

How many times area is changed, when sides of a triangle are doubled. -Maths 9th

Description : How many times area is changed, when sides of a triangle are doubled. -Maths 9th

Last Answer : Four times area is changed, when sides of a triangle are doubled.

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

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

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

Last Answer : Answer:- As we know herons formula = √[s(s-a)(s-b)(s-c)] Now here a=15cm, b=8cm, c=17cm And, s=20cm Then, area is=√[20*5*12*3] =√3600 =60cm² Therefore the Area is 60cm². Regards, Rishikesh

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.

If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

Description : If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

Last Answer : answer:

ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th

Description : ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th

Last Answer : answer:

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Description : 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Last Answer : Solution: Given that, Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O. To prove that, The Quadrilateral ABCD is a square. Proof, In ΔAOB and ΔCOD, AO = ... right angle. Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square. Hence Proved.

4. Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Description : 4. Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Last Answer : Solution: Let ABCD be a square and its diagonals AC and BD intersect each other at O. To show that, AC = BD AO = OC and ∠AOB = 90° Proof, In ΔABC and ΔBAD, AB = BA (Common) ∠ABC = ∠BAD = ... = ∠COB ∠AOB+∠COB = 180° (Linear pair) Thus, ∠AOB = ∠COB = 90° , Diagonals bisect each other at right angles

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Description : Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Last Answer : Solution :-

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Description : Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Last Answer : We have a quadrilateral ABCD such that angleO is the mid-point of AC and BD. Also AC ⊥ BD. Now, in ΔAOD and ΔAOB, we have AO = AO [Common] OD = OB [ ... i.e. The rhombus ABCD is having one angle equal to 90°. Thus, ABCD is a square.

Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Description : Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Last Answer : Proof: (i) In a ΔABC and ΔBAD, AB=AB ( common line) BC=AD ( opppsite sides of a square) ∠ABC=∠BAD ( = 90° ) ΔABC≅ΔBAD( By SAS property) AC=BD ( by CPCT). (ii) In a ΔOAD and ΔOCB, ... ) ( by CPCT ∠AOB+∠AOD=180° (linear pair) 2∠AOB=180° ∠AOB=∠AOD=90° ∴AC and BD bisect each other at right angles.

Can a triangle have two obtuse angles? Give reason for your answer. -Maths 9th

Description : Can a triangle have two obtuse angles? Give reason for your answer. -Maths 9th

Last Answer : No, because if the triangle have two obtuse angles i.e., more than 90° angle, then the sum of all three angles of a triangle will not be equal to 180°.

Prove that a triangle must have at least two acute angles. -Maths 9th

Description : Prove that a triangle must have at least two acute angles. -Maths 9th

Last Answer : Given ΔABC is a triangle. To prove ΔABC must have two acute angles Proof Let us consider the following cases Case I When two angles are 90°. Suppose two angles are ∠B = 90° and ∠C = 90°

Can a triangle have two obtuse angles? Give reason for your answer. -Maths 9th

Description : Can a triangle have two obtuse angles? Give reason for your answer. -Maths 9th

Last Answer : No, because if the triangle have two obtuse angles i.e., more than 90° angle, then the sum of all three angles of a triangle will not be equal to 180°.