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Triangle. Ratio of the area of triangles. Theorem 1. Example In the figure, BC// DE, AC= 3 cm and CE= 4 cm. Find. Class work In the figure, PSQ, QXR and RYP are straight lines. If the area of is , find the area of the parallelogram SXRY.

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## Triangle

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**Ratio of the area of triangles**Theorem 1**Example**In the figure, BC// DE, AC= 3 cm and CE= 4 cm. Find**Class work**In the figure, PSQ, QXR and RYP are straight lines. If the area of is , find the area of the parallelogram SXRY.**81**36 Area of parallelogram = 225 – 81 - 36 = 108 Area of parallelogram is**Triangle**C height A B base**Triangle**base height**Triangle**base height**Area of Triangle**C height A B base**For triangles with commonheight,**Theorem 2**Eg.3) Given that BC : DC = 5: 1**Find and have common height,**Eg.3) Given that BC : DC = 5: 1**Find and have common height, = 4**Eg.4) Given that AD = 3 cm and CD = 1 cm.**Find and have common height,**Class work**5.) In the figure, find the area of : area of .**Class work**5.) In the figure, find the area of : area of .**Class work**5.) In the figure, find the area of : area of . = 1**6.) In the figure,**given that QX:XR = 5:6 and PY:YR =5:3 calculate the area of (a) (b)**6.) In the figure,**given that QX:XR = 5:6 and PY:YR =5:3 5x 6x**6.) In the figure,**given that QX:XR = 5:6 and PY:YR =5:3 5x 6x**6.) In the figure,**given that QX:XR = 5:6 and PY:YR =5:3 5y 3y 5x 6x**6.) In the figure,**given that QX:XR = 5:6 and PY:YR =5:3 5y 3y 5x 6x**7) In the figure, PQRS is a rectangle.**M is a midpoint of QR. PR and MS intersects at N. Find the area of : area of PQMN.**In the figure, PQRS is a rectangle.**M is a midpoint of QR. PR and MS intersects at N. Find the area of : area of PQMN.**In the figure, PQRS is a rectangle.**M is a midpoint of QR. PR and MS intersects at N. Find the area of : area of PQMN.**Find the area of : area of PQMN.**4A A = 4**Find the area of : area of PQMN.**4A 2y 2A y A Considering They have the common height.**Find the area of : area of PQMN.**4A 2y 2A y A Hence, = 2A : 5A = 2 : 5**8.) In the figure, PX:XQ = 1: 2, PY:YR = 3:2.**Area of : Area of = ?**In the figure, PX:XQ = 1: 2, PY:YR = 3:2.**Area of : Area of = ? and have the common height**In the figure, PX:XQ = 1: 2, PY:YR = 3:2.**Area of : Area of = ? and have the common height**In the figure, PX:XQ = 1: 2, PY:YR = 3:2.**Area of : Area of = ? and have the common height = 2A : 5A = 2 : 5**9.) In the figure, PQRS is a rectangle. RSX is a straight**line and PX// QS. If the area of PQRS is 24 and Y is a point on QR such that QY :YR = 3:1, find the area of .**10.) In the figure, if**, Find**10.) In the figure, if**, A 3A**10.) In the figure, if**, A x 3A 3x**10.) In the figure, if**, A x 3A 3x 9A**10.) In the figure, if**, A x 3A 3A 3x 9A**10.) In the figure, if**, A x 3A 3A 3x 9A**Ratio of the area of**similar triangles Theorem 1**For triangles with commonheight,**Theorem 2

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