Question

10th class math geometry
practice set 1.1
qo no.1
base of a triangle is 9 and height is 5 best of another triangle is 10 and height is a 6 find the ratio of area of this triangle.​

Answer 1

Heya !!

Here's your answer.. ⬇⬇

♦ Given :- Let The triangle be ∆ABC and ∆PQR.

In ∆ABC,

Base of ∆ABC is BC = 9cm

Altitude of ∆ABC is AE = 5cm

In ∆PQR,

Base is QR = 10cm

Altitude is PM = 6cm

♦ To Find :- Ratio of Area of ∆ABC and ∆PQR

♦ Solution :-

$\begin {gathered} = \frac{area \: \: of \: \: abc}{area \: \: of \: \: pqr} \\ \\ = \frac{ \frac{1}{2} \times ae \times bc }{ \frac{1}{2} \times pm \times qr } \\ \\ = \frac{ \frac{1}{2} \times 5 \times 9 }{ \frac{1}{2} \times 6 \times 10} \\ \\ = \frac{3}{4} \\ \\ \end{gathered}$

Hence, Ratio of Area of ∆ABC to ∆PQR is 3:4

Answer 2

Answer:

Let ABC and PQR be two right triangles with AB ⊥ BC and PQ ⊥ QR.

Given: BC = 9, AB = 5, PQ = 6 and QR = 10.

AreaABC

AreaPQR

ACBC

PQQR

Area(△ABC)Area(△PQR)=12×AC×BC12×PQ×QR

=

12×5×912×6×10

=

34

Hence, Ratio of Area of △ABC : Area of △PQR =

3: 4.