Question

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 triangle.​

Answer 1

Answer:

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

Answer 2

Answer:

ANSWER

Given :-

First Triangle :-

Height = 5 cm

Base = 9 cm

$area = \frac{1}{2} \times b \times h \\ \frac{1}{2} \times 9 \times 5 = \frac{45}{2} = 22.5 {cm}^{2}$

Second triangle : -

Height = 6 cm

Base = 10 cm

$\frac{1}{2} \times 10 \times 6 = 5 \times 6 \\ = 30 {cm}^{2}$

Therefore:-

Final answers

22.5 : 30

ANSWER

4.5 : 6

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