9th Grade > Mathematics
HERON S FORMULA MCQs
Total Questions : 51
| Page 1 of 6 pages
:
The area of an equilateral triangle is √34a2, where a = 2 cm.
Hence the area is √34×22
=√3cm2
Thus the value of A = 3.
Answer: Option B. -> 12 cm2
:
B
Since ABCD is a parallelogram, its opposite sides will be equal.
AB = 3 cm, BC =4 cm and AC = 5 cm
Now in triangle ABC,
Area of the triangle using heron’s formula,
s=Perimeter2=3+4+52=6cm
Area =√s(s−a)(s−b)(s−c)
=√6(6−3)(6−4)(6−5)
= √36
= 6cm2
Area of the parallelogram = 2 (Area of triangle ABC)
⇒The area of the parallelogram=2×6=12cm2
:
B
Since ABCD is a parallelogram, its opposite sides will be equal.
AB = 3 cm, BC =4 cm and AC = 5 cm
Now in triangle ABC,
Area of the triangle using heron’s formula,
s=Perimeter2=3+4+52=6cm
Area =√s(s−a)(s−b)(s−c)
=√6(6−3)(6−4)(6−5)
= √36
= 6cm2
Area of the parallelogram = 2 (Area of triangle ABC)
⇒The area of the parallelogram=2×6=12cm2
Answer: Option B. -> False
:
B
The given statement is false. Heron's formula can be used in finding the area of quadrilateralas quadrilateral is formed by two triangles.
:
B
The given statement is false. Heron's formula can be used in finding the area of quadrilateralas quadrilateral is formed by two triangles.
Answer: Option B. -> False
:
B
Area of a triangle=12×base×height=40
⇒base=40×2height
⇒base=8010=8cm
:
B
Area of a triangle=12×base×height=40
⇒base=40×2height
⇒base=8010=8cm
Answer: Option A. -> True
:
A
In a right angled triangle, if the base and the height are given, the area can be calculated directly by using 12 × base × height. If the hypotenuse and either the base or the height are given, then the third side can be found using Pythagoras' theorem. Then the area can be calculated using the same formula.
:
A
In a right angled triangle, if the base and the height are given, the area can be calculated directly by using 12 × base × height. If the hypotenuse and either the base or the height are given, then the third side can be found using Pythagoras' theorem. Then the area can be calculated using the same formula.
Answer: Option C. -> 4 cm
:
C
Let the side of the rhombus = a
Area of the triangle containing the diagonal = 12 times area of the rhombus
One of the diagonal = 6 cm
Semi perimeter = 6+a+a2=a+3
Now the area of the triangle containing the diagonal =√s(s−a)(s−b)(s−c)
3√7 = √(a+3)(a+3−a)(a+3−a)(a+3−6)
=√(a+3)(a−3)(9)
= √(a2−9)(9)
We take square on both sides and get,
63 = (a2−9)×9
⇒(a2−9) = 7
⇒a2 = 7+9 = 16
⇒a= 4cm
:
C
Let the side of the rhombus = a
Area of the triangle containing the diagonal = 12 times area of the rhombus
One of the diagonal = 6 cm
Semi perimeter = 6+a+a2=a+3
Now the area of the triangle containing the diagonal =√s(s−a)(s−b)(s−c)
3√7 = √(a+3)(a+3−a)(a+3−a)(a+3−6)
=√(a+3)(a−3)(9)
= √(a2−9)(9)
We take square on both sides and get,
63 = (a2−9)×9
⇒(a2−9) = 7
⇒a2 = 7+9 = 16
⇒a= 4cm
Answer: Option C. -> 16√5
:
C
Side of the rhombus = 6 cm
One of the diagonal = 8 cm
Semi perimeter of the triangle containing the diagonal = (6+6+8)2 = 10 cm
Area of the triangle containing the diagonal =√s(s−a)(s−b)(s−c)
= √10(10−6)(10−6)(10−8)
= √10×4×4×2
= √2×5×4×4×2
⇒8√5cm2
Area of rhombus = 2 (Area of the triangle containing the diagonal)
Area of the rhombus = 2×8√5 = 16√5cm2
:
C
Side of the rhombus = 6 cm
One of the diagonal = 8 cm
Semi perimeter of the triangle containing the diagonal = (6+6+8)2 = 10 cm
Area of the triangle containing the diagonal =√s(s−a)(s−b)(s−c)
= √10(10−6)(10−6)(10−8)
= √10×4×4×2
= √2×5×4×4×2
⇒8√5cm2
Area of rhombus = 2 (Area of the triangle containing the diagonal)
Area of the rhombus = 2×8√5 = 16√5cm2
Answer: Option C. -> 12 cm2
:
C
Let each of its equal sides be x.
Then, 2x+8=18 ⇒x=5cm
Hence, the sides are 5cm, 5cm and 8 cm.
Using the Heron's formula,
Area(A)=√s(s−a)(s−b)(s−c)
where s = semi- perimeter and a,b,c are the three sides of the triangle.
Semi−Perimeter(s)=perimeter2
=182 =9cm
A=√9(9−5)(9−5)(9−8)
=√9×4×4×1
=√144cm2
=12cm2
:
C
Let each of its equal sides be x.
Then, 2x+8=18 ⇒x=5cm
Hence, the sides are 5cm, 5cm and 8 cm.
Using the Heron's formula,
Area(A)=√s(s−a)(s−b)(s−c)
where s = semi- perimeter and a,b,c are the three sides of the triangle.
Semi−Perimeter(s)=perimeter2
=182 =9cm
A=√9(9−5)(9−5)(9−8)
=√9×4×4×1
=√144cm2
=12cm2
:
(Area of triangle) =12×base×height
The three sides are (10cm, 6cm and 8cm).
102=62+82
The sides satisfy Pythagoras theorem. Hence, this is a right-angled triangle, with the sides 6cm and 8cm being the base and the height, respectively.
Hence, Area=12×6×8 cm2
=24cm2
Answer: Option A. -> 9√3 cm
:
A
Area of square = Side2
⇒36 =Side2
⇒ Side = 6 cm
Side of triangle = 6 cm
Semi perimeter of the triangle s =6+6+62cm=9cm
Area of the triangle =√s(s−a)(s−b)(s−c)
=√9(9−6)3
=√243
=9√3cm
:
A
Area of square = Side2
⇒36 =Side2
⇒ Side = 6 cm
Side of triangle = 6 cm
Semi perimeter of the triangle s =6+6+62cm=9cm
Area of the triangle =√s(s−a)(s−b)(s−c)
=√9(9−6)3
=√243
=9√3cm
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