In-centre of a triangle lies in the interior of -Maths 9th

In-centre of a triangle lies in the interior of -Maths 9th
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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

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

Last Answer : each interior opposite angles are 55

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

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

Last Answer : each interior opposite angles are 55

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Description : From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Last Answer : Let each side of ㎝ equilateral triangle ABC be ′a′㎝ Now, ar△OAB=21 AB OP=21 a 14=7a㎠→1 ar△OBC= BC OQ =21 a 10=5a㎠→2 ar△OAC=21 AC OR=21 a 6=3a㎠→3 ∴ar△ABC=1+2+3=7a+5a+3a=15a㎠ Also area of equilateral ... ABC=43 a2 Now, 43 a2=15a⇒a=3 15 4 3 3 =3603 =203 ㎝ Now, ar△ABC=43 (203 )2=3003 ㎠

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Description : From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Last Answer : Area of triangle =

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

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle.Show that △OCD is an isosceles triangle. -Maths 9th

Description : O is a point in the interior of a square ABCD such that OAB is an equilateral triangle.Show that △OCD is an isosceles triangle. -Maths 9th

Last Answer : Solution :-

The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

Description : The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

Last Answer : (c) (5, 0) or (1, 4) Let the third vertex of the triangle be P(a, b). Since it lies on the line x + y = 5, a + b = 5 ...(i) Also, given area of triangle formed by the points (2, -1), (3, 2) and (a, b) = 4 ... b) - (-3a + b) = 5 + 15⇒ 4a = 20 ⇒ a = 5 ⇒ b = 0. ∴ The points are (1, 4) and (5, 0).

If O is the centre of the circle and chord AB = OA and the area of triangle AOB -Maths 9th

Description : If O is the centre of the circle and chord AB = OA and the area of triangle AOB -Maths 9th

Last Answer : (b) 16π cm2.AB = OA ⇒ AB = OA = OB (radii of circle are equal) ⇒ ΔAOB is equilateral. ∴ If ‘r’ is the radius of the circle,then area of ΔAOB = \(rac{\sqrt3}{4}\)side2⇒ \(rac{\sqrt3}{4}\)(r)2 = 4√3 (given)⇒ r2 = 16 ⇒ r = 4∴ Area of circle = πr2 = 16π cm2.

Find the co-ordinates of the in-centre of the triangle whose vertices are (–36, 7), (20, 7) and (0, –8). -Maths 9th

Description : Find the co-ordinates of the in-centre of the triangle whose vertices are (–36, 7), (20, 7) and (0, –8). -Maths 9th

Last Answer : Let A(1, 2), B(0, -1) and C(2, -1) be the mid-points of the sides PQ, QR and RP of the triangle PQR. Let the co-ordinates of P, Q and R be (x1, y1), (x2, y2) and (x3 , y3) respectively. Then, by the mid- ... ordinates of centroid of ΔPQR = \(\bigg(rac{3+(-1)+1}{3},rac{2+2+(-4)}{3}\bigg)\) = (1, 0).

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Description : In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Last Answer : Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles] ⇒ 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Description : In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Last Answer : Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles] ⇒ 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Description : A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Last Answer : Solution :- The two lines will not be always parallel as the sum of the two equal angles will not always be 180°. Lines will be parallel when each of the equal angles is equal to 90°.

Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Description : Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Last Answer : Solution :-

The lengths of the perpendiculars drawn from any point in the interior of an equilateral -Maths 9th

Description : The lengths of the perpendiculars drawn from any point in the interior of an equilateral -Maths 9th

Last Answer : (a) \(rac{2}{\sqrt3}\) (p1 + p2 + p3) Let each side of equilateral ΔPQR = a units. O is any point in the interior of DΔPQR ⇒ OD = p1, OE = p2 and OF = p3 are perpendiculars on sides PQ, PR and QR respectively. ∴ Area of ... }{4}a^2\) = \(rac{a}{2}(p_1+p_2+p_3)\) ⇒ \(rac{2}{\sqrt3}\) (p1 + p2 + p3)

From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th

Description : From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th

Last Answer : (i) In Δ O C E ,D C 2 = D E 2 + E C 2 Δ O B D , D B 2 = O D 2 + B D 2 Δ O A F , O A 2 = O F 2 + A F 2 Adding we get O A 2 + O B 2 + O C 2 = O F 2 + O D 2 + O F 2 + E C 2 + B D 2 + A F 2 A F 2 + B D 2 + C E 2 = O A

A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Then, show that OA^2 + OC^2 = OB^2 + OD^2. -Maths 9th

Description : A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Then, show that OA^2 + OC^2 = OB^2 + OD^2. -Maths 9th

Last Answer : answer:

If two parallel lines are intersected by a transversal, then the bisectors of the interior angles form which one of the following? -Maths 9th

Description : If two parallel lines are intersected by a transversal, then the bisectors of the interior angles form which one of the following? -Maths 9th

Last Answer : answer:

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

Name the quadrant in which the point lies : -Maths 9th

Description : Name the quadrant in which the point lies : -Maths 9th

Last Answer : (a) A (1, 1) lies in the 1st quadrant, (b) B (2, 4) lies in the 1st quadrant. (c) C (-3, -10) lies in the 3rd quadrant. (d) D (-1, 2) lies in the 2nd quadrant. (e) E (1 , -1) lies in the ... lies in the 3rd quadrant. (g) G (-3, 10) lies in the 2nd quadrant. (h) H (1, -2) lies in the 4th quadrant.

Point (-3, 5) lies in the -Maths 9th

Description : Point (-3, 5) lies in the -Maths 9th

Last Answer : (b) In point (-3, 5), x-coordinate is negative and y-coordinate is positive. So, the point lies in the second quadrant.

Point (0, – 7) lies -Maths 9th

Description : Point (0, – 7) lies -Maths 9th

Last Answer : (c) In point (0, -7) x-coordinate is zero, so it lies on Y-axis and y-coordinate is negative, so the point (0, – 7) lies on the Y-axis in the negative direction.

Point (- 10,0) lies -Maths 9th

Description : Point (- 10,0) lies -Maths 9th

Last Answer : (a) In point (-10, 0) y-coordinate is zero, so it lies on X-axis and its x-coordinate is negative, so the point (-10, 0) lies on the X-axis in the negative direction.

If y-coordinate of a point is zero, then this point always lies -Maths 9th

Description : If y-coordinate of a point is zero, then this point always lies -Maths 9th

Last Answer : (c) If y-coordinate of a point is zero, then this point always lies on X-axis. Because perpendicular distance of the point from X-axis measured along Y-axis is zero.

If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Description : The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Last Answer : (b) Given ordinate of the point is 4 arid the point lies on Y-axis, so its abscissa is zero. Hence, the required point is (0, 4).

The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Description : The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Last Answer : (C) Given the point lies on X-axis this shows that its ^-coordinate is zero. Also, it is at a distance of 5 units in negative direction of X-axis, so its y-coordinate” is negative.Hence, the required point is (0, – 5).

Which of the following points lies on Y-axis ? -Maths 9th

Description : Which of the following points lies on Y-axis ? -Maths 9th

Last Answer : We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also ... 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

The point of the form (a, a) always lies on -Maths 9th

Description : The point of the form (a, a) always lies on -Maths 9th

Last Answer : (c) Since, the given point (a, a) has same value of x and y-coordinates. Therefore, the point (a, a), must be lie on the line y = x.

The point of the form (a, – a) always lies on the line -Maths 9th

Description : The point of the form (a, – a) always lies on the line -Maths 9th

Last Answer : (d) Taking option (d), x + y = a + (-a) = a – a = 0 [since, give point is of the form (a, -a)] Hence, the point (a, – a) always lies on the line x + y = 0.

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Description : If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Last Answer : Since, the point (x = 3, y = 4) lies on the equation 3y = ax + 7, then the equation will be , satisfied by the point. Now, put x = 3 and y = 4 in given equation, we get 3(4) = a (3)+7 ⇒ 12 = 3a+7 ⇒ 3a = 12 – 7 ⇒ 3a = 5 Hence, the value of a is 5/3.

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

Name the quadrant in which the point lies : -Maths 9th

Description : Name the quadrant in which the point lies : -Maths 9th

Last Answer : (a) A (1, 1) lies in the 1st quadrant, (b) B (2, 4) lies in the 1st quadrant. (c) C (-3, -10) lies in the 3rd quadrant. (d) D (-1, 2) lies in the 2nd quadrant. (e) E (1 , -1) lies in the ... lies in the 3rd quadrant. (g) G (-3, 10) lies in the 2nd quadrant. (h) H (1, -2) lies in the 4th quadrant.

Point (-3, 5) lies in the -Maths 9th

Description : Point (-3, 5) lies in the -Maths 9th

Last Answer : (b) In point (-3, 5), x-coordinate is negative and y-coordinate is positive. So, the point lies in the second quadrant.

Point (0, – 7) lies -Maths 9th

Description : Point (0, – 7) lies -Maths 9th

Last Answer : (c) In point (0, -7) x-coordinate is zero, so it lies on Y-axis and y-coordinate is negative, so the point (0, – 7) lies on the Y-axis in the negative direction.

Point (- 10,0) lies -Maths 9th

Description : Point (- 10,0) lies -Maths 9th

Last Answer : (a) In point (-10, 0) y-coordinate is zero, so it lies on X-axis and its x-coordinate is negative, so the point (-10, 0) lies on the X-axis in the negative direction.

If y-coordinate of a point is zero, then this point always lies -Maths 9th

Description : If y-coordinate of a point is zero, then this point always lies -Maths 9th

Last Answer : (c) If y-coordinate of a point is zero, then this point always lies on X-axis. Because perpendicular distance of the point from X-axis measured along Y-axis is zero.

If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Description : The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Last Answer : (b) Given ordinate of the point is 4 arid the point lies on Y-axis, so its abscissa is zero. Hence, the required point is (0, 4).

The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Description : The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Last Answer : (C) Given the point lies on X-axis this shows that its ^-coordinate is zero. Also, it is at a distance of 5 units in negative direction of X-axis, so its y-coordinate” is negative.Hence, the required point is (0, – 5).

Which of the following points lies on Y-axis ? -Maths 9th

Description : Which of the following points lies on Y-axis ? -Maths 9th

Last Answer : We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also ... 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

The point of the form (a, a) always lies on -Maths 9th

Description : The point of the form (a, a) always lies on -Maths 9th

Last Answer : (c) Since, the given point (a, a) has same value of x and y-coordinates. Therefore, the point (a, a), must be lie on the line y = x.

The point of the form (a, – a) always lies on the line -Maths 9th

Description : The point of the form (a, – a) always lies on the line -Maths 9th

Last Answer : (d) Taking option (d), x + y = a + (-a) = a – a = 0 [since, give point is of the form (a, -a)] Hence, the point (a, – a) always lies on the line x + y = 0.

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Description : If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Last Answer : Since, the point (x = 3, y = 4) lies on the equation 3y = ax + 7, then the equation will be , satisfied by the point. Now, put x = 3 and y = 4 in given equation, we get 3(4) = a (3)+7 ⇒ 12 = 3a+7 ⇒ 3a = 12 – 7 ⇒ 3a = 5 Hence, the value of a is 5/3.

if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : Since complete line is AC and B is point on it. therefore, AC is divide into 2 parts AB&BC. therefore, AC=AB+BC

if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : AB=AC-BC BC =AC-AB AB+BC=AB HENCE PROVED

Check whether the point (a ,– a) lies on y=x–a or not. -Maths 9th

Description : Check whether the point (a ,– a) lies on y=x–a or not. -Maths 9th

Last Answer : Solution :-