CBSE Questions for Class 9 Maths Areas Of Parallelograms And Triangles Quiz 1 - MCQExams.com

$${\textbf{Step - 1: Find the length of PQ and PR}}$$

                  $${\text{In right angled triangle OPQ ,we have}}$$

                  $${\text{PQ = }}\sqrt {{{(OP)}^2} - {{(OQ)}^2}} $$

                  $$ = \sqrt {169 - 25}  = \sqrt {144}  = 12\ cm$$

                  $$ \Rightarrow PR = 12\ cm\ \ \     {\text{(Two tangents from the same external point to a circle are equal)}}$$

$${\textbf{Step - 2: Find Area of quadrilateral}}$$ 

                  $${\text{ = (2}} \times \dfrac{1}{2} \times 12 \times 5)\ c{m^2} = 60\ c{m^2}$$              $$\textbf{[area of PQOR = 2} \times {\textbf{Area of }}\Delta {\textbf{POQ}]}$$

$${\textbf{Hence , area of the required quadrilateral is (A) 60 c}}{{\textbf{m}}^2}$$

$$\textbf{Step -1: Find the area of required triangle and parallelogram.}$$

                $$\text{As, the }\triangle ABD\text{ and parallelogram }ABCD\text{ are on same base and between same parallels.}$$

              $$\therefore \text{Height of the triangle and parallelogram is same.}$$

              $$\text{Area of }\triangle ABD=\dfrac{1}{2}\times\text{Base}\times\text{Height}$$

              $$=\dfrac{1}{2}\times AB\times DE$$

              $$\text{Area of parallelogram}=\text{Base}\times\text{Height}$$

              $$=AB\times DE$$

$$\textbf{Step -2: Find the required ratio.}$$

                $$\dfrac{\text{Area of }\triangle ABD}{\text{Area of parallelogram }ABCD}=\dfrac{\dfrac{1}{2}\times AB\times DE}{AB\times DE}$$

                $$=\dfrac{1}{2}$$

$$\textbf{Hence, option B is correct.}$$