Mix Examples - Heron’s Formula Questions in English

(D) Join $B$ and $D$ to divide the quadrilateral $ABCD$ into two triangles: $\Delta BCD$ and $\Delta ABD$.
$1$. In the right-angled $\Delta BCD$ (since $\angle C = 90^{\circ}$):
Area of $\Delta BCD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 5 = 30 \, m^{2}$.
$2$. Calculate the length of the diagonal $BD$ using the Pythagoras theorem:
$BD^{2} = BC^{2} + CD^{2} = 12^{2} + 5^{2} = 144 + 25 = 169$.
$BD = \sqrt{169} = 13 \, m$.
$3$. For $\Delta ABD$,the sides are $a = 9 \, m$,$b = 8 \, m$,and $c = 13 \, m$.
Semi-perimeter $s = \frac{9 + 8 + 13}{2} = \frac{30}{2} = 15 \, m$.
Using Heron's formula,Area of $\Delta ABD = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{15(15-9)(15-8)(15-13)} = \sqrt{15 \times 6 \times 7 \times 2} = \sqrt{1260} = 6\sqrt{35} \, m^{2}$.
Using $\sqrt{35} \approx 5.916$,Area $\approx 6 \times 5.916 = 35.496 \approx 35.5 \, m^{2}$.
$4$. Total area of quadrilateral $ABCD = \text{Area}(\Delta BCD) + \text{Area}(\Delta ABD) = 30 + 35.5 = 65.5 \, m^{2}$.

(B) Let the two equal sides of the isosceles right triangle be $a \, cm$.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Since it is an isosceles right triangle,base $= a$ and height $= a$.
Therefore,$\frac{1}{2} \times a \times a = 8$.
$a^{2} = 16$,which gives $a = 4 \, cm$.
By the Pythagorean theorem,the hypotenuse $h$ is given by $h = \sqrt{a^{2} + a^{2}} = \sqrt{4^{2} + 4^{2}} = \sqrt{16 + 16} = \sqrt{32} \, cm$.

(B) The area of a triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Given,$\text{base} = 4 \, \text{cm}$ and $\text{height} = 6 \, \text{cm}$.
Substituting these values into the formula: $\text{Area} = \frac{1}{2} \times 4 \, \text{cm} \times 6 \, \text{cm} = 12 \, \text{cm}^2$.
Since the calculated area is $12 \, \text{cm}^2$ and not $24 \, \text{cm}^2$,the statement is False.

(TRUE) Given that $\Delta ABC$ is a right-angled triangle with $\angle A = 90^{\circ}$,$AB = 4 \, cm$,and $AC = 4 \, cm$.
The area of a right-angled triangle is given by the formula:
Area $= \frac{1}{2} \times \text{base} \times \text{height}$
Here,we can take $AC$ as the base and $AB$ as the height.
Area $= \frac{1}{2} \times 4 \, cm \times 4 \, cm$
Area $= \frac{1}{2} \times 16 \, cm^2 = 8 \, cm^2$
Since the calculated area is $8 \, cm^2$,the given statement is True.

(A) Let $ABCD$ be the rhombus whose one diagonal $AC$ is $16 \, cm$. Each side of the rhombus is $10 \, cm$.
We know that the diagonals of a rhombus bisect each other at right angles,so
$OA = OC = 8 \, cm$ and $OB = OD$.
In $\Delta AOB$,we have $\angle AOB = 90^{\circ}$.
Therefore,$AB^2 = OA^2 + OB^2$
$\Rightarrow OB^2 = AB^2 - OA^2$
$= (10)^2 - 8^2 = 100 - 64 = 36$
$OB = \sqrt{36} = 6 \, cm$.
$DB = 2(OB) = 2 \times 6 = 12 \, cm$.
Hence,the area of the rhombus $= \frac{1}{2} \times \text{Product of diagonals}$
$= \frac{1}{2} \times 16 \times 12 = 96 \, cm^2$.
Hence,the given statement is true.

(FALSE) The statement is False.
A regular hexagon can be divided into six equilateral triangles by joining its center to each of its vertices.
Since each of these six triangles is an equilateral triangle with side $a$, the total area of the regular hexagon is equal to the sum of the areas of these six equilateral triangles.
Therefore, the area of a regular hexagon of side $a$ is $6 \times (\text{Area of an equilateral triangle of side } a)$, not the sum of the areas of five such triangles.

(N/A) The figure consists of two triangles. The shaded area is the difference between the area of the larger triangle and the smaller triangle.
$1$. For the larger triangle with sides $a = 122 \ m$,$b = 120 \ m$,and $c = 22 \ m$:
Semi-perimeter $s = \frac{122 + 120 + 22}{2} = \frac{264}{2} = 132 \ m$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{132(132-122)(132-120)(132-22)}$
$= \sqrt{132 \times 10 \times 12 \times 110} = \sqrt{1742400} = 1320 \ m^2$.
$2$. For the smaller triangle with sides $a = 24 \ m$,$b = 26 \ m$,and $c = 22 \ m$:
Since it is a right-angled triangle (as indicated in the figure),we can use the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.
Area $= \frac{1}{2} \times 24 \times 10 = 120 \ m^2$.
(Note: Using Heron's formula also yields the same result: $s = \frac{24+26+22}{2} = 36$. Area $= \sqrt{36(36-24)(36-26)(36-22)} = \sqrt{36 \times 12 \times 10 \times 14} = \sqrt{60480} \approx 245.9 \ m^2$. However,given the right angle symbol,the base-height method is more accurate).
$3$. Area of the shaded region $= \text{Area of larger triangle} - \text{Area of smaller triangle}$
$= 1320 \ m^2 - 120 \ m^2 = 1200 \ m^2$.

(N/A) The sides of $\Delta BCD$ are $a = 12 \text{ cm}, b = 17 \text{ cm},$ and $c = 25 \text{ cm}.$
$\therefore$ The semi-perimeter $(s)$ of $\Delta BCD$ is:
$s = \frac{a + b + c}{2} = \frac{12 + 17 + 25}{2} = \frac{54}{2} = 27 \text{ cm}.$
$\therefore$ Area of $\Delta BCD = \sqrt{s(s - a)(s - b)(s - c)}$ [By Heron's formula]
$= \sqrt{27(27 - 12)(27 - 17)(27 - 25)}$
$= \sqrt{27 \times 15 \times 10 \times 2}$
$= \sqrt{(9 \times 3) \times (3 \times 5) \times (5 \times 2) \times 2}$
$= \sqrt{9 \times 9 \times 25 \times 4} = 3 \times 3 \times 5 \times 2 = 90 \text{ cm}^2.$
Since a diagonal divides a parallelogram into two triangles of equal area:
Area of parallelogram $ABCD = 2 \times \text{Area of } \Delta BCD = 2 \times 90 = 180 \text{ cm}^2.$
Let the altitude from vertex $A$ to side $DC$ be $h.$
Area of parallelogram $ABCD = \text{Base} \times \text{Altitude}$
$180 = DC \times h$
$180 = 12 \times h$
$h = \frac{180}{12} = 15 \text{ cm}.$
Hence,the area of the parallelogram is $180 \text{ cm}^2$ and the length of the altitude is $15 \text{ cm}.$

(B) To find the area of quadrilateral $ABCD$,we divide it into two triangles: $\Delta ABC$ and $\Delta ACD$.
Since $\angle B = 90^\circ$,$\Delta ABC$ is a right-angled triangle.
By Pythagoras theorem in $\Delta ABC$:
$AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm$.
Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 8 = 24 \, cm^2$.
Now,for $\Delta ACD$,the sides are $a = 10 \, cm$,$b = 12 \, cm$,and $c = 14 \, cm$.
The semi-perimeter $s = \frac{a + b + c}{2} = \frac{10 + 12 + 14}{2} = \frac{36}{2} = 18 \, cm$.
Using Heron's formula,Area of $\Delta ACD = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18(18-10)(18-12)(18-14)} = \sqrt{18 \times 8 \times 6 \times 4} = \sqrt{3456} \approx 58.787 \, cm^2$.
Total area of quadrilateral $ABCD = \text{Area}(\Delta ABC) + \text{Area}(\Delta ACD) = 24 + 58.787 = 82.787 \, cm^2$.

(C) The perimeter of the rhombus is $40 \, cm$.
Since all sides of a rhombus are equal,the side length is $\frac{40}{4} = 10 \, cm$.
The diagonal divides the rhombus into two congruent triangles with sides $10 \, cm, 10 \, cm,$ and $12 \, cm$.
Using Heron's formula for one triangle:
Semi-perimeter $s = \frac{10 + 10 + 12}{2} = \frac{32}{2} = 16 \, cm$.
Area of one triangle $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{16(16-10)(16-10)(16-12)} = \sqrt{16 \times 6 \times 6 \times 4} = \sqrt{2304} = 48 \, cm^2$.
Area of the rhombus $= 2 \times 48 = 96 \, cm^2$.
Since both sides are painted,the total area to be painted is $2 \times 96 = 192 \, cm^2$.
The cost of painting at the rate of $Rs \, 5$ per $cm^2$ is $192 \times 5 = Rs \, 960$.

(A) Draw $RT \perp SP$. From the figure,it is clear that $RT = PQ = 7 \ m$ and $PT = QR = 7 \ m$.
Since $SP = 12 \ m$,we have $ST = SP - PT = 12 \ m - 7 \ m = 5 \ m$.
Now,in the right-angled triangle $STR$,by Pythagoras theorem:
$SR^2 = ST^2 + RT^2$
$13^2 = 5^2 + RT^2$
$169 = 25 + RT^2$
$RT^2 = 169 - 25 = 144$
$RT = 12 \ m$.
Thus,the height of the trapezium is $12 \ m$.
The area of trapezium $PQRS = \frac{1}{2} \times (SP + QR) \times PQ$
$= \frac{1}{2} \times (12 \ m + 7 \ m) \times 12 \ m$
$= \frac{1}{2} \times 19 \ m \times 12 \ m = 19 \times 6 \ m^2 = 114 \ m^2$.

(A) Let the trapezium be $ABCD$ with parallel sides $AB = 90 \, m$ and $CD = 30 \, m$. The side $AC$ is perpendicular to both $AB$ and $CD$.
Draw $DM \perp AB$. Since $ACDM$ forms a rectangle,$AM = CD = 30 \, m$ and $DM = AC$.
Now,$MB = AB - AM = 90 \, m - 30 \, m = 60 \, m$.
In the right-angled triangle $\triangle DMB$,by Pythagoras theorem:
$DM^2 = DB^2 - MB^2 = (100)^2 - (60)^2$
$DM^2 = 10000 - 3600 = 6400$
$DM = \sqrt{6400} = 80 \, m$.
Thus,the height of the trapezium is $80 \, m$.
The area of the trapezium $ABCD = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$= \frac{1}{2} \times (90 + 30) \times 80 = \frac{1}{2} \times 120 \times 80 = 4800 \, m^2$.
Total cost of ploughing = $\text{Area} \times \text{Rate} = 4800 \times 4 = \operatorname{Rs} 19,200$.

(N/A) $ABCD$ is a rectangle in which $AB = 51 \, cm$ and $BC = 25 \, cm$.
Since the parallel sides $QC$ and $PD$ are in the ratio $9:8$,let $QC = 9x$ and $PD = 8x$.
The height of the trapezium $PQCD$ is the same as the width of the rectangle,which is $25 \, cm$.
The area of the trapezium $PQCD = \frac{1}{2} \times (QC + PD) \times \text{height} = \frac{1}{2} \times (9x + 8x) \times 25 = \frac{1}{2} \times 17x \times 25$.
The area of the rectangle $ABCD = \text{length} \times \text{width} = 51 \times 25$.
It is given that the area of the trapezium $PQCD = \frac{5}{6} \times \text{Area of rectangle } ABCD$.
Therefore,$\frac{1}{2} \times 17x \times 25 = \frac{5}{6} \times 51 \times 25$.
Dividing both sides by $25$ and multiplying by $2$,we get $17x = \frac{5}{6} \times 51 \times 2 = \frac{5}{3} \times 51 = 5 \times 17 = 85$.
Thus,$x = \frac{85}{17} = 5$.
Hence,the length $QC = 9x = 9 \times 5 = 45 \, cm$.
And the length $PD = 8x = 8 \times 5 = 40 \, cm$.

(D) regular hexagon consists of $6$ equilateral triangles with side length $a = 12\,cm$.
Perimeter of a regular hexagon $= 6 \times a = 6 \times 12\,cm = 72\,cm$.
Area of one equilateral triangle $= \frac{\sqrt{3}}{4} \times a^2 = \frac{\sqrt{3}}{4} \times 12^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3}\,cm^2$.
Total area of the hexagon $= 6 \times 36\sqrt{3}\,cm^2 = 216\sqrt{3}\,cm^2$.
Thus,the perimeter is $72\,cm$ and the area is $216\sqrt{3}\,cm^2$.

(D) The diagonal $BD$ of quadrilateral $ABCD$ divides it into two non-overlapping triangles,$\Delta ABD$ and $\Delta BCD$.
For $\Delta ABD$,the sides are $a = 40 \, cm$,$b = 29 \, cm$,and $c = 29 \, cm$.
The semi-perimeter $s_1 = \frac{40 + 29 + 29}{2} = \frac{98}{2} = 49 \, cm$.
Area of $\Delta ABD = \sqrt{s_1(s_1 - a)(s_1 - b)(s_1 - c)} = \sqrt{49(49 - 40)(49 - 29)(49 - 29)} = \sqrt{49 \times 9 \times 20 \times 20} = 7 \times 3 \times 20 = 420 \, cm^2$.
For $\Delta BCD$,the sides are $a = 29 \, cm$,$b = 48 \, cm$,and $c = 35 \, cm$.
The semi-perimeter $s_2 = \frac{29 + 48 + 35}{2} = \frac{112}{2} = 56 \, cm$.
Area of $\Delta BCD = \sqrt{s_2(s_2 - a)(s_2 - b)(s_2 - c)} = \sqrt{56(56 - 29)(56 - 48)(56 - 35)} = \sqrt{56 \times 27 \times 8 \times 21} = \sqrt{(8 \times 7) \times (9 \times 3) \times 8 \times (3 \times 7)} = \sqrt{8^2 \times 7^2 \times 3^2 \times 9} = 8 \times 7 \times 3 \times 3 = 504 \, cm^2$.
Area of quadrilateral $ABCD = \text{Area of } \Delta ABD + \text{Area of } \Delta BCD = 420 + 504 = 924 \, cm^2$.

(B) Let $ABCD$ be the plot which is divided by the diagonal $BD = 72 \, m$ into two equal triangles.
Since $ABCD$ is a rhombus,all its sides are equal.
Perimeter $= 4 \times \text{side} = 340 \, m$.
Therefore,side $= \frac{340}{4} = 85 \, m$.
In $\Delta ABD$,the sides are $a = 85 \, m$,$b = 85 \, m$,and $c = 72 \, m$.
Semi-perimeter $(s) = \frac{85 + 85 + 72}{2} = \frac{242}{2} = 121 \, m$.
Using Heron's formula,Area of $\Delta ABD = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{121(121-85)(121-85)(121-72)}$
$= \sqrt{121 \times 36 \times 36 \times 49}$
$= 11 \times 36 \times 7 = 2772 \, m^2$.
Since the diagonal divides the rhombus into two equal triangles,the total area of the rhombus $ABCD = 2 \times \text{Area of } \Delta ABD$
$= 2 \times 2772 = 5544 \, m^2$.

(A) $1$. Divide the quadrilateral $ABCD$ into two triangles by drawing diagonal $AC$.
$2$. In $\triangle ABC$,since $\angle B = 90^{\circ}$,it is a right-angled triangle.
$3$. Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 15 = 150 \, cm^2$.
$4$. Using Pythagoras theorem in $\triangle ABC$,$AC^2 = AB^2 + BC^2 = 20^2 + 15^2 = 400 + 225 = 625$. Thus,$AC = 25 \, cm$.
$5$. Now,in $\triangle ADC$,the sides are $a = 25 \, cm$,$b = 12 \, cm$,$c = 17 \, cm$. The semi-perimeter $s = \frac{25 + 12 + 17}{2} = \frac{54}{2} = 27 \, cm$.
$6$. Area of $\triangle ADC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27(27-25)(27-12)(27-17)} = \sqrt{27 \times 2 \times 15 \times 10} = \sqrt{8100} = 90 \, cm^2$.
$7$. Total area of quadrilateral $ABCD = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 150 + 90 = 240 \, cm^2$.

(D) The quadrilateral $ABCD$ is divided into two triangles,$\triangle ABC$ and $\triangle ADC$,by the diagonal $AC = 48 \, cm$.
For $\triangle ABC$,the sides are $a = 29 \, cm$,$b = 35 \, cm$,and $c = 48 \, cm$.
The semi-perimeter $s_1 = (29 + 35 + 48) / 2 = 112 / 2 = 56 \, cm$.
Area of $\triangle ABC = \sqrt{s_1(s_1-a)(s_1-b)(s_1-c)} = \sqrt{56(56-29)(56-35)(56-48)} = \sqrt{56 \times 27 \times 21 \times 8} = \sqrt{254016} = 504 \, cm^2$.
For $\triangle ADC$,the sides are $a = 50 \, cm$,$b = 14 \, cm$,and $c = 48 \, cm$.
The semi-perimeter $s_2 = (50 + 14 + 48) / 2 = 112 / 2 = 56 \, cm$.
Area of $\triangle ADC = \sqrt{s_2(s_2-a)(s_2-b)(s_2-c)} = \sqrt{56(56-50)(56-14)(56-48)} = \sqrt{56 \times 6 \times 42 \times 8} = \sqrt{112896} = 336 \, cm^2$.
Total area of quadrilateral $ABCD = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 504 + 336 = 840 \, cm^2$.

(A) $1$. Join $AC$ to divide the quadrilateral $ABCD$ into two triangles: $\triangle ABC$ and $\triangle ADC$.
$2$. In right-angled $\triangle ABC$,by Pythagoras theorem: $AC^2 = AB^2 + BC^2 = 9^2 + 40^2 = 81 + 1600 = 1681$. Thus,$AC = \sqrt{1681} = 41 \, cm$.
$3$. Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 40 \times 9 = 180 \, cm^2$.
$4$. For $\triangle ADC$,the sides are $a = 41 \, cm$,$b = 28 \, cm$,and $c = 15 \, cm$. Semi-perimeter $s = \frac{41 + 28 + 15}{2} = \frac{84}{2} = 42 \, cm$.
$5$. Area of $\triangle ADC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{42(42-41)(42-28)(42-15)} = \sqrt{42 \times 1 \times 14 \times 27} = \sqrt{(14 \times 3) \times 14 \times (9 \times 3)} = \sqrt{14^2 \times 3^2 \times 3^2} = 14 \times 3 \times 3 = 126 \, cm^2$.
$6$. Total area of quadrilateral $ABCD = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 180 + 126 = 306 \, cm^2$.

(A) To find the area of the trapezium $PQRS$,draw a line $ST$ perpendicular to $PQ$ at $T$.
Since $PQ \parallel SR$ and $QR \perp PQ$,$QR \perp SR$ as well. Thus,$TQRS$ is a rectangle.
Given $PQ = 12 \, m$ and $SR = 7 \, m$,we have $TQ = SR = 7 \, m$.
Therefore,$PT = PQ - TQ = 12 \, m - 7 \, m = 5 \, m$.
In the right-angled triangle $\triangle PTS$,by Pythagoras theorem:
$TS^2 = PS^2 - PT^2 = 13^2 - 5^2 = 169 - 25 = 144$.
So,$TS = \sqrt{144} = 12 \, m$.
Since $TQRS$ is a rectangle,$QR = TS = 12 \, m$.
The area of the trapezium is given by the formula:
$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$\text{Area} = \frac{1}{2} \times (PQ + SR) \times QR$
$\text{Area} = \frac{1}{2} \times (12 + 7) \times 12 = 19 \times 6 = 114 \, m^2$.