Manipur Board Solutions for Chapter: Triangles, Exercise 4: EXERCISE 7.4

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Manipur Board Mathematics Solutions for Exercise - Manipur Board Solutions for Chapter: Triangles, Exercise 4: EXERCISE 7.4

Attempt the practice questions on Chapter 7: Triangles, Exercise 4: EXERCISE 7.4 with hints and solutions to strengthen your understanding. Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from Manipur Board Solutions for Chapter: Triangles, Exercise 4: EXERCISE 7.4 with Hints & Solutions

EASY
10th Manipur Board
IMPORTANT

ABC is a right triangle right-angled at B and D is the foot of the perpendicular drawn from B on AC. If DMBC and DNAB where M, N lie on BC, AB respectively, prove that DM2=DN×MC.

EASY
10th Manipur Board
IMPORTANT

ABC is a right triangle right-angled at B and D is the foot of the perpendicular drawn from B on AC. If DMBC and DNAB where M, N lie on BC, AB respectively, prove that DN2=DM×AN.

HARD
10th Manipur Board
IMPORTANT

Through the vertex D of parallelogram ABCD, a line is drawn to intersects the sides BA produced and BC produced at E and F respectively. Prove that ADAE=BFBE=CFCD

EASY
10th Manipur Board
IMPORTANT

The perimeter of two similar triangle ABC and PQR respectively, 72 cm and 48 cm. If PQ=20 cm, find AB

EASY
10th Manipur Board
IMPORTANT

Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and  BD intersects at PP, prove that AP×PC=PD×PB 

EASY
10th Manipur Board
IMPORTANT

If one angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite sides in the same ratio, prove that the triangles are similar.

EASY
10th Manipur Board
IMPORTANT

If two triangles are equiangular, prove that the ratio of the corresponding sides is same as the ratio of the corresponding altitudes.

EASY
10th Manipur Board
IMPORTANT

If two triangles are equiangular , prove that the ratio of the corresponding sides is same as the ratio of the corresponding angle bisector segments.