Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find theratio of areas of these triangles.
Answers
Step-by-step explanation:
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}=areaofpqrareaofabc=21×pm×qr21×ae×bc=21×6×1021×5×9=43
Hence, Ratio of Area of ∆ABC to ∆PQR is 3:4
(please make me brainlist)
Answer:
3:4
Step-by-step explanation:
The area of a triangle is 1/2bh
Area of triangle 1: 1/2*9*5
= 45/2 sq. units
Area of triangle 2: 1/2*10*6
=60/2sq.units
Ratio
45/2:60/2
45:60
3:4