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Syllabus

10. In the figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC.

IF BH is perpendicular is FG, prove that :

(ii) Area of the square ABDE = Area of the rectangle ARHF.

9.In the given figure, AP is parallel to BC, BP is parallel to CQ. Prove that the areas of triangles ABC and BQP are equal.no 12 a)

Q12.(a) In the figure (1) given below, ABCD and AEFG are two parallelograms. Prove that area of || gm ABCD = area of || gm AEFG.

Q.6. In the figure given below, ABC is an equilateral triangle in which P and S are the mid-points of arcs AB and AC respectively. Prove that: PQ = QR = RS.

In the figure given below, AB || DC || EF, AD || BC and ED || FA. Prove that: Area of DEFH = Area of ABCD

Q.5. The base of a right angled triangle is 12 cm and its hypotenuse is 13 cm long. Find its area and the perimeter.

10. ABC is a triangle in which AB=AC=4 cm and $\angle \mathrm{A}=90\xb0$. Calculate the area of $\u25b3$ABC. Also find the length of perpendicular from A and BC.

Hint: By Pythagoras theorem, BC

^{2}=AB^{2}+AC^{2}=4^{2}+4^{2}=32 $\Rightarrow $BC =4$\sqrt{2}$cm.Q.13. The sides of a right-angled triangle containing the right angle are 5x cm and (3x - 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 $c{m}^{2}$.

Q.15. If the perimeter of a right angled triangle is 60 cm and its hypotenuse is 25 cm, find its area.

Q.14. In $\u25b3$ABC, $\angle B=90\xb0$, AB = (2x + 1) cm and BC = (x + 1) cm. If the area of the $\u25b3$ABC is 60 $c{m}^{2}$, find its perimeter.

Q.9 (i) If the lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 48 cm, find its area.

No link plzz as i am not able to acess it

RTP:ar(AOB)=ar(AOD)

Const:Join BD

Q10. In the given figure, AP is parallel to BC, BP is parallel to CQ. Prove that areas of triangles ABC and BQP are equal.

Q.12. Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.

(i) area of ∆ ACD = area of ∆ ABE

(ii) area of ∆ OBD = area of ∆ OCE