Question

Base of a triangle is 6 and height is 4. Base of another triangle is 5 and height is 3. Find the ratio of areas of these triangles​

Answer 1

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}

=

areaofpqr

areaofabc

=

2

1

×pm×qr

2

1

×ae×bc

=

2

1

×6×10

2

1

×5×9

=

4

3

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

Hope it helps..

Explanation:

Hope it helps army

Borahae ( ◜‿◝ )♡

And if numbers are not same just follow the process you will get your answer :)

Answer 2

Let the first triangle be ∆ABC

Let the second triangle be ∆DEF

In ∆ABC :-

Base b = 6; Height h = 4

Area A = ½ × b × h

A = ½ × 6 × 4

A = 1 × 3 × 4

A = 12

In ∆DEF :-

Base b = 5; Height h = 3

Area A = ½ × 5 × 3

A = 15/2

Ratio of Area of ∆ABC : Area of ∆DEF :-

Area of ∆ABC : Area of ∆DEF

12 : 15/2

8 : 5