ΔMNT ~ ΔQRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio

Solve Question:

The ratio of the areas of two triangles with the common base is 4:3. Height of the larger triangle is 2 cm, then finds the corresponding height of the smaller triangle.

Solve any five sub-questions:

ΔDEF ∼ ΔMNK. If DE = 5, MN = 6, then find the value of

Solve any two sub-questions:

Prove that “That ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.”

In fig 1.80, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.

Activity :

.......... by componendo.

.......... (I)

.......... test of similarity.

.......... corresponding sides of similar triangles.

...from (I)

In ΔPQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.

Complete the proof by filling in the boxes. In ΔPMQ, ray MX is bisector of ∠PMQ.

.......... (I) theorem of angle bisector.

In ΔPMR, ray MY is bisector of ∠PMR.

.......... (II) theorem of angle bisector.

But .......... M is the midpoint QR, hence MQ = MR.

∴ XY || QR .......... converse of basic proportionality theorem.

Solve sub-questions:

In Δ ABC, B - D – C and BD = 7, BC = 20 then find following ratios.

(1)

(2)

(3)

In □ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that

If ΔABC ~ ΔPQR, A (ΔABC) = 80, A(ΔPQR) = 125, then fill in the blanks.

In figure 1.81, the vertices of square DEFG are on the sides of ΔABC, ∠A = 90°. Then prove that DE² = BD × EC

(Hint : Show that ΔGBD is similar to ΔDFE. Use GD = FE = DE.)