**Question 1.Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.Solution:**

Let the base, height and area of the first triangle be b

_{1}, h

_{1}, and A

_{1}respectively.

Let the base, height and area of the second triangle be b

_{2}, h

_{2}and A

_{2}respectively.

[Since Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]

∴ The ratio of areas of the triangles is 3:4.

**Question 2.Solution:**

∆ABC and ∆ADB have same base AB.

[Since Triangles having equal base]

**Question 3.In the adjoining figure, seg PS ± seg RQ, seg QT ± seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.Solution:**

**Question 4.In the adjoining figure, AP ⊥ BC, AD || BC, then find A(∆ABC) : A(∆BCD).Solution:**

Draw DQ ⊥ BC, B-C-Q.

AD || BC [Given]

∴ AP = DQ (i) [Perpendicular distance between two parallel lines is the same]

∆ABC and ∆BCD have same base BC.

**Question 5.In the adjoining figure, PQ ⊥ BC, AD ⊥ BC, then find following ratios.Solution:**

i. ∆PQB and tPBC have same height PQ.

ii. ∆PBC and ∆ABC have same base BC.

iii. ∆ABC and ∆ADC have same height AD.

**Question 1.Solution:**

In ∆ABC, BC is the base and AR is the height.

In ∆APQ, PQ is the base and AR is the height.