Dhondoop's field has adjesent angles in the ratio 4:5. Find all the angles of his field. -Maths 9th

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Description : Dhondoop's field has adjesent angles in the ratio 4:5. Find all the angles of his field. -Maths 9th

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Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle.

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle.

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle

Description : Angles of a triangle are in the ratio 2:4:3. The smallest angle of the triangle is -Maths 9th

Last Answer : (b) Given, the ratio of angles of a triangle is 2 : 4 : 3. Let the angles of a triangle be ∠A, ∠B and ∠C. ∠A = 2x, ∠B = 4x ∠C = 3x , ∠A+∠B+ ∠C= 180° [sum of all the angles of a triangle is 180°] 2x ... ∠B = 4x = 4 x 20° = 80° ∠C = 3x = 3 x 20° = 60° Hence, the smallest angle of a triangle is 40°.

Description : If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a -Maths 9th

Last Answer : (c) Given, ratio of angles of quadrilateral ABCD is 3 : 7 : 6 : 4. Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively. We know that, sum of all angles of a quadrilateral is 360°. 3x + 7x + 6x + 4x = 360° => 20x = 360° => x=360°/20° = 18°

Description : Angles of a triangle are in the ratio 2:4:3. The smallest angle of the triangle is -Maths 9th

Last Answer : (b) Given, the ratio of angles of a triangle is 2 : 4 : 3. Let the angles of a triangle be ∠A, ∠B and ∠C. ∠A = 2x, ∠B = 4x ∠C = 3x , ∠A+∠B+ ∠C= 180° [sum of all the angles of a triangle is 180°] 2x ... ∠B = 4x = 4 x 20° = 80° ∠C = 3x = 3 x 20° = 60° Hence, the smallest angle of a triangle is 40°.

Description : If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a -Maths 9th

Last Answer : (c) Given, ratio of angles of quadrilateral ABCD is 3 : 7 : 6 : 4. Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively. We know that, sum of all angles of a quadrilateral is 360°. 3x + 7x + 6x + 4x = 360° => 20x = 360° => x=360°/20° = 18°

Description : If the ratio between two complementary angles is 2: 3, then find the angles. -Maths 9th

Last Answer : 2x+3x=180 5x=180 X=36

Description : The angles of a triangle are in the ratio 3: 7: 8. Find the angles of the triangle. -Maths 9th

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Description : Two adjacent angles of a ||gm are in the ratio 2:3. Find all the four angles of the parallelogram. -Maths 9th

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Description : The angles of a quadrilateral are in the ratio 1:2:3:4. Find all the angles of the quadrilateral. -Maths 9th

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Description : If the angles of a triangle are in the ratio 1 : 2 : 3, then find the ratio of the corresponding opposite sides. -Maths 9th

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Description : The angles of a triangle are in the ratio 8 : 3 : 1. What is the ratio of the longest side of the triangle to the next longest side? -Maths 9th

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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.

Description : The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Last Answer : Let sides of △ are=6x,7x and 8x Perimeter=6x+7x+8x=21x 21x=410 x=20 Sides are 120,140 and 160 m Area =S(S−A)(S−B)(S−C)​ [Heron's Formula] S=2120+140+160​=210 m A=210(210−120)(210−140)(210−160)​=210015​ sq. m

Description : The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Last Answer : Let sides of △ are=6x,7x and 8x Perimeter=6x+7x+8x=21x 21x=410 x=20 Sides are 120,140 and 160 m Area =S(S−A)(S−B)(S−C)​ [Heron's Formula] S=2120+140+160​=210 m A=210(210−120)(210−140)(210−160)​=210015​ sq. m

Description : While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Last Answer : (a) Yes , x < y is correct (b) Ð ADB =Ð DBC = y (alternate int. angles) since BC < CD (angle opp. to smaller side is smaller) there for, x < y (c) Truth value

Description : 2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. -Maths 9th

Last Answer : Consider the following diagram- Here, it is given that AOB = COD i.e. they are equal angles. Now, we will have to prove that the line segments AB and CD are equal i.e. AB = CD. Proof: In triangles AOB ... ) So, by SAS congruency, ΔAOB ΔCOD. ∴ By the rule of CPCT, we have AB = CD. (Hence proved).

Description : 1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. -Maths 9th

Last Answer : To recall, a circle is a collection of points whose every point is equidistant from its centre. So, two circles can be congruent only when the distance of every point of both the circles are equal from the centre ... ) So, by SSS congruency, ΔAOB ΔCOD ∴ By CPCT we have, AOB = COD. (Hence proved).

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

Description : 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral whose diagonals bisect each other at right angles. Given that, OA = OC OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90° To show that, if the ... a parallelogram. , ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle. Hence Proved.

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

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

Description : If one of a parallelogram is twice of its adjacent angle , find the angles of the parallelogram . -Maths 9th

Last Answer : Let the two adjacent angles be x° and 2x° . In a parallelogram, sum of the adjacent angles are 180°. ∴ x + 2x = 180° ⇒ 3x = 180° ⇒ x = 60° Thus , the two adjacent angles are 120° and 60°. Hence, the angles of the parallelogram are 120°, 60°, 120° and 60°.

Description : If the diagonals of a quadrilateral bisect each other at right angles , then name the quadrilateral . -Maths 9th

Last Answer : Quadrilateral will be Rhombus .

Description : ABCD is a parallelogram and line segments AX, CY bisect the angles A and C, respectively. -Maths 9th

Last Answer : Since opposite angles are equal in a parallelogram . Therefore , in parallelogram ABCD , we have ∠A = ∠C ⇒ 1 / 2 ∠A = 1 / 2 ∠C ⇒ ∠1 = ∠2 ---- i) [∵ AX and CY are bisectors of ∠A and ∠C ... intersects AX and YC at A and Y such that ∠1 = ∠3 i.e. corresponding angles are equal . ∴ AX | | CY .

Description : In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th

Last Answer : Given : A parallelogram ABCD such that the bisectors of adjacent angles A and B intersect at P. To prove : ∠APB = 90° Proof : Since ABCD is a | | gm ∴ AD | | BC ⇒ ∠A + ∠B = 180° [sum of consecutive interior ... 90° + ∠APB + ∠2 = 180° [ ∵ ∠1 + ∠2 = 90° from (i)] Hence, ∠APB = 90°

Description : In the given figure , two opposite angles of a parallelograms PQRS are (3x - 4)° and (56 - 3x)° . Find all the angles of given parallelogram . -Maths 9th

Last Answer : We know that opposite angles of a paralleloram are equal . ∴ ∠P = ∠R ⇒ 3x - 4 = 56 - 3x ⇒ 6x = 60 ⇒ x = 10 Thus, ∠P = ∠R = 3 10 - 4 = 26° Also, ∠P + ∠Q = 180° [ ... Hence, the four angles of the parallelogram PQRS are 26°, 154°,26° and 154°. We should care our earth to have good eco balance.

Description : Equal chords of a circle subtend equal angles at the centre. -Maths 9th

Last Answer : Given : In a circle C(O,r), chord AB = chord CD. To Prove : ∠AOB = ∠COD. Proof : In△AOB and △COD AO = CO [radii of same circle] BO = DO [radii of same circle] Chord AB = Chord CD [given] ⇒ △AOB ≅ △COD [by SSS congruence axiom] ⇒ ∠AOB = ∠COD. [c.p.c.t.]

Description : If the angles subtended by the chords of a circle at the centre are equal, then chords are equal. -Maths 9th

Last Answer : Given : In a circle C(O,r) , ∠AOB = ∠COD To Prove : Chord AB = Chord CD . Proof : In △AOB and △COD AO = CO [radii of same circle] BO = DO [radii of same circle] ∠AOB = ∠COD [given] ⇒ △AOB ≅ △COD [by SAS congruence axiom] ⇒ Chord AB = Chord CD [c.p.c.t]

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

Last Answer : (d) Let the angles of a AABC be ∠A, ∠B and ∠C. Given, ∠A = ∠B+∠C …(i) InMBC, ∠A+ ∠B+ ∠C-180° [sum of all angles of a triangle is 180°]…(ii) From Eqs. (i) and (ii), ∠A+∠A = 180° ⇒ 2 ∠A = 180° ⇒ 180° /2 ∠A = 90° Hence, the triangle is a right triangle.

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°.

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.

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]

Description : If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

Last Answer : Given Two lines AB and CD intersect at point O.

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°

Description : Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Last Answer : Solution of this question

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

Description : All the angles of a quadrilateral are equal. What special name is given to this quadrilateral ? -Maths 9th

Last Answer : We know that, sum of all angles in a quadrilateral is 360°. If ABCD is a quadrilateral, ∠A+ ∠B+ ∠C + ∠D = 360° (i) But it is given all angles are equal. ∠A = ∠B = ∠C = ∠D From Eq. (i ... ⇒ 4 ∠A = 360° ∠A = 90° So, all angles of a quadrilateral are 90°. Hence, given quadrilateral is a rectangle.

Description : Can all the four angles of a quadrilateral be obtuse angles? Give reason for your answer. -Maths 9th

Last Answer : No, all the four angles of a quadrilateral cannot be obtuse. As, the sum of the angles of a quadrilateral is 360°, then may have maximum of three obtuse angles.

Description : Can all the angles of a quadrilateral be acute angles ? Give reason for your answer. -Maths 9th

Last Answer : No, all the angles of a quadrilateral cannot be acute angles. As, sum of the angles of a quadrilateral is 360°. So, maximum of three acute angles will be possible.

Description : Can all the angles of a quadrilateral be right angles? Give reason for your answer. -Maths 9th

Last Answer : Yes, all the angles of a quadrilateral can be right angles. In this case, the quadrilateral becomes rectangle or square.

Description : Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Last Answer : Given, opposite angles of a quadrilateral are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. ∴ CD = AB = 4cm

Description : ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

Last Answer : According to question altitude from D to side AB bisects AB. Find the angles of the rhombus.

Description : A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Last Answer : According to question parallelogram bisects one of its angles.