CBSE Questions for Class 9 Maths Heron'S Formula Quiz 1 - MCQExams.com

The sides of a triangle are $$3$$ cm, $$4$$ cm and $$5$$ cm. Its area is .......
  • $$12 cm^2$$
  • $$15 cm^2$$
  • $$20 cm^2$$
  • $$6 cm^2$$
State true or false:
In $$\triangle ABC$$, $$AB= BC=CA=2a$$ and AD is perpendicular to side BC, the area of $$\triangle ABC= a^2\sqrt2$$.
  • True
  • False
Area of an equilateral triangle of side $$a$$ units can be calculated by using the formula :
  • $$\sqrt{s^{2}(s-a)^{2}}$$
  • $$(s-a)\sqrt{s^{2}(s-a)}$$
  • $$\sqrt{s(s-a)^{2}}$$
  • $$(s-a)\sqrt{s(s-a)}$$
The area of an equilateral triangle with side $$2 \sqrt{3}$$ cm is
  • $$5.196 cm^2$$
  • $$0.866 cm^2$$
  • $$3.496 cm^2$$
  • $$1.732 cm^2$$
The area of a triangle whose sides are 4 cm, 13 cm and 15 cm is
  • $$ \displaystyle \sqrt{420}$$ $$ \displaystyle cm^{2}$$
  • 24 $$ \displaystyle cm^{2} $$
  • 48 $$ \displaystyle cm^{2} $$
  • 56 $$ \displaystyle cm^{2} $$
The sides of a triangle are $$4, 5$$ and $$6$$ cm. The area of the triangle is equal to
  • $$\displaystyle \frac{15}{4} cm^2$$
  • $$\displaystyle \frac{15}{4}\sqrt 7 cm^2$$
  • $$\displaystyle \frac{4}{15} \sqrt 7 cm^2$$
  • None of these
If the sides of triangle are doubled then is area
  • remains the same
  • becomes doubled
  • becomes three times
  • become four times
Heron's formula is :
  • $$\Delta = \sqrt{s(s+a)(s+b)(s+c)}$$
  • $$\Delta = \sqrt{(s-a)(s-b)(s-c)}$$
  • $$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$, $$s = a+b+c$$
  • $$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$, $$2s = a+b+c$$
Area of traingle ABC whose sides are 24m, 40m and 32m is-
  • $$96{m}^{2}$$
  • $$384{m}^{2}$$
  • $$43{m}^{2}$$
  • $$192{m}^{2}$$
The sides of a triangle are 5 cm, 12 cm and 13 cm. Then its area  is
  • $$0.0024 m^2$$
  • $$0.0026 m^2$$
  • $$0.003 m^2$$
  • $$0.0015 m^2$$
If s is the semi-perimeter of a $$\Delta A B C$$ whose sides are a,b,c then s=.............?
  • $$a + b + c$$
  • $$\dfrac { a + b + c } { 2 }$$
  • $$\dfrac { a + b + c } { 3 }$$
  • $$\dfrac { a + b + c } { 4 }$$
Find the area of a triangle whose two sides are $$18$$ cm and $$10$$ cm and the perimeter is $$42$$ cm.
  • $$83.67$$ cm$$^2$$
  • $$46.12$$ cm $$^2$$
  • $$77.15$$ cm$$^2$$
  • $$69.69$$ cm$$^2$$
The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is
  • $$1322 cm^2$$
  • $$1311 cm^2$$
  • $$1344 cm^2$$
  • $$1392 cm^2$$
Find the cost of laying grass in a triangular field of sides $$50\>m, 65\>m$$ and $$65\> m$$ at the rate of Rs $$7$$ per $$m^2$$.
  • Rs $$10500$$
  • Rs $$11375$$
  • Rs $$10000$$
  • Rs $$21000$$
The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is
  • $$\sqrt{15} cm^2$$
  • $$\sqrt{\dfrac{15}{2}}cm^2$$
  • $$2 \sqrt{15} cm^2$$
  • $$4 \sqrt{15}cm^2$$
In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude corresponding to the side having length 12 cm is 
  • 10.25 cm 
  • 11.25 cm 
  • 12.25 cm 
  • 9.25 cm 
The sides of a $$\Delta$$ are $$7 cm, 24 cm$$ and $$25 cm$$. Find its area.
  • $$168 cm^2$$
  • $$84 cm^2$$
  • $$87.5 cm^2$$
  • $$300 cm^2$$
The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of Rs $$3$$ per $$m^2$$ is 
  • Rs. 918
  • Rs. 818
  • Rs. 718
  • Rs. 618
Area of triangle ABC whose sides are 24 m, 40 m and 32 m, is -
  • $$90 m^2$$
  • $$384 m^2$$
  • $$43 m^2$$
  • $$192 m^2$$
The area of an isosceles triangle whose base is $$'b'$$ and equal sides are of length $$'a'$$ is:
  • $$\dfrac{a}{4}\sqrt{4b^2-a^2}$$
  • $$\dfrac{b}{4}\sqrt{4a^2-b^2}$$
  • $$\dfrac{b}{2}\sqrt{4b^2-a^2}$$
  • $$\dfrac{a}{2}\sqrt{4a^2-b^2}$$
The sides of a triangular plot are in the ratio $$4 : 5 : 6$$ and its perimeter is $$150\ \mathrm{cm}$$. Then the sides are:
  • $$4\ \mathrm{cm}, 5\ \mathrm{cm}, 6\ \mathrm{cm}$$
  • $$40\ \mathrm{ cm}, 50\ \mathrm{ cm}, 60\ \mathrm{ cm}$$
  • $$8\ \mathrm{cm}, 10\ \mathrm{cm}, 12\ \mathrm{cm}$$
  • $$120\ \mathrm{cm}, 150\ \mathrm{cm}, 180\ \mathrm{cm}$$
Two sides of a triangle are $$13 cm$$ and $$14 cm$$ and its semi perimeter is $$18 cm$$. Then third side of the triangle is :
  • $$12 cm$$
  • $$11 cm$$
  • $$10 cm$$
  • $$9 cm$$
In figure, $$Ar(||gm ABCD)$$ is:
84416.png
  • $$10 cm$$
  • $$20 cm$$
  • $$10\sqrt{3}cm^2$$
  • $$20\sqrt{3}cm^2$$
Two sides of a triangle are $$16 cm$$ and $$14 cm$$ and its semi-perimeter is $$18 cm$$. Then the third side of the triangle is:
  • $$12 cm$$
  • $$11 cm$$
  • $$10 cm$$
  • $$6 cm$$
Area of an equilateral triangle of side 'a' units can be calculated by using the formula :
  • $$\sqrt{s^2(s-a)^2}$$
  • $$(s-a)\sqrt{s^2(s-a)}$$
  • $$\sqrt{s(s-a)^2}$$
  • $$(s-a)\sqrt{s(s-a)}$$
Heron's formunla is :
  • $$\Delta = \sqrt{s(s+a)(s+b)(s+c)}$$
  • $$\Delta = \sqrt{(s-a)(s-b)(s-c)}$$
  • $$\Delta = \sqrt{s(s-a)(s-b)(s-c)} , s = a+b+c$$
  • $$\Delta = \sqrt{s(s-a)(s-b)(s-c)} , 2s = a+b+c$$
Heron's formula is :
  • $$\Delta =\sqrt{s(s+a)(s+b)(s+c)}$$
  • $$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$$
  • $$\Delta =\sqrt{s(s-a)(s-b)(s-c)}, s=a+b+c$$
  • $$\Delta =\sqrt{s(s-a)(s-b)(s-c)}, 2s=a+b+c$$
The area of a triangle whose sides are $$13$$ cm, $$14$$ cm and $$15$$ cm is :
  • $$42 ~cm^{2}$$
  • $$86~cm^{2}$$
  • $$84~cm^{2}$$
  • $$100~cm^{2}$$
Area of an equilateral triangle of side a units can be calculated by using the formula :
  • $$\sqrt{s^2(s-a)^2}$$
  • $$(s-a)\sqrt{s^2(s-a)}$$
  • $$\sqrt{s(s-a)^3}$$
  • $$(s-a)\sqrt{s(s-a)}$$
The lengths of four sides and a diagonal of the given  quadrilateral are indicated in  the diagram. If $$A$$ denotes the area and $$l$$ the length of the other diagonal, then $$A$$ and $$l$$ are respectively

97698_46cef8cee5814e5dbdc5296c359fb2a0.png
  • $$12\sqrt{6},4\sqrt{6}$$
  • $$12\sqrt{6},5\sqrt{6}$$
  • $$6\sqrt{6},4\sqrt{6}$$
  • $$6\sqrt{6},5\sqrt{6}$$
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