Question

27) The base and height of a triangle ABC is 6 and 7. The base and height of another triangle PQR is 14 and 8 then A(AABC): A(APQR) * O 6:14 O 3:8 08:14 O 7:8​

Answer 1

QUESTION-:

The base and height of a triangle ABC is 6 and 7. The base and height of another triangle PQR is 14 and 8 then ArAABC): Ar(APQR)

a)6:14

b) 3:8

c)8:14

d) 7:8

EXPLANATION-:

We know that -:

$\longrightarrow \underline{\boxed{\tt \; \dag\;Area_{(triangle)}=\frac{1}{2}\;base \times height }}$

∴Area of ΔABC-:

$\rightarrow \bf\; Area_{(\triangle ABC)}=\frac{1}{2}\times 6\times 7\\\\\rightarrow \bf\; Area_{(\triangle ABC)}= 3\times 7\\\\\\\longrightarrow \boxed{\bf\; Area_{(\triangle ABC)}=21\; unit^2}$

& Area of ΔPQR-:

​$\rightarrow \bf\; Area_{(\triangle PQR)}=\frac{1}{2}\times 14\times 8\\\\\rightarrow \bf\; Area_{(\triangle PQR)}= 7\times 8\\\\\\\longrightarrow \boxed{\bf\; Area_{(\triangle PQR)}=56\; unit^2}$

Now finding ratio-:

$\dashrightarrow \bf\; \frac{Ar(\triangle ABC)}{Ar(\triangle PQR )} =\frac{21\;unit^2}{56\;unit^2}\\\\\dashrightarrow \bf\; \frac{Ar(\triangle ABC)}{Ar(\triangle PQR )} =\frac{3}{8}$

∴Ratio of Ar(AABC): Ar(APQR) is 3:8

Answer 2

Answer:

your answer is attached above