A rhombus shaped field has green grass for 18 cows to graze.
[question] Question. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? [/question] [solution] Solution: Let ABCD be a rhombus-shaped field. For $\triangle B C D$, Semi-perimeter, $s=\frac{(48+30+30) \mathrm{cm}}{2}=54 \mathrm{~m}$ By Heron’s formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Therefore, area of $\triangle B C D=[\sqrt{54(54-48)(54-30)(54-30)}] \...
Read More →A triangle and a parallelogram have the same base and the same area
[question] Question A triangle and a parallelogram have the same base and the same area. If the sides of triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram. [/question] [solution] Solution: For triangle Perimeter of triangle $=(26+28+30) \mathrm{cm}=84 \mathrm{~cm}$ $2 s=84 \mathrm{~cm}$ $s=42 \mathrm{~cm}$ By Heron's formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Area of triangle $=[\sqrt{42(42-26)(42-28)(42-30)}] \mat...
Read More →Radha made a picture of an aeroplane with coloured papers as shown in the given figure.
[question] Question. Radha made a picture of an aeroplane with coloured papers as shown in the given figure. Find the total area of the paper used. [solution] Solution: For triangle I This triangle is an isosceles triangle. Perimeter $=2 s=(5+5+1) \mathrm{cm}=11 \mathrm{~cm}$ $s=\frac{11 \mathrm{~cm}}{2}=5.5 \mathrm{~cm}$ Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=[\sqrt{5.5(5.5-5)(5.5-5)(5.5-1)}] \mathrm{cm}^{2}$ $=[\sqrt{(5.5)(0.5)(0.5)(4.5)}] \mathrm{cm}^{2}$ $=0.75 \sqrt{11} \mathrm{~...
Read More →Find the area of a quadrilateral ABCD in which AB = 3 cm
[question] Question. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. [/question] [solution] Solution: For $\triangle \mathrm{ABC}$, $A C^{2}=A B^{2}+B C^{2}$ $(5)^{2}=(3)^{2}+(4)^{2}$ Therefore, $\triangle \mathrm{ABC}$ is a right-angled triangle, right-angled at point $\mathrm{B}$. Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}=\frac{1}{2} \times 3 \times 4=6 \mathrm{~cm}^{2}$ For $\triangle \mathrm{ADC}...
Read More →A park, in the shape of a quadrilateral ABCD,
[question] Question. A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy? [/question] [solution] Solution: Let us join BD. In $\triangle B C D$, applying Pythagoras theorem, $B D^{2}=B C^{2}+C D^{2}$ $=(12)^{2}+(5)^{2}$ $=144+25$ $B D^{2}=169$ $B D=13 \mathrm{~m}$ Area of $\triangle B C D=\frac{1}{2} \times B C \times C D=\left(\frac{1}{2} \times 12 \times 5\right) \mathrm{m}^{2}=30 \mathrm{~m}^{2}$ For $\triangle \...
Read More →Convert the following temperature to Celsius scale :
[question] Question Convert the following temperature to Celsius scale : (a) 300 K (b) 573 K [/question] [solution] Solution (a) 300 – 273 = 27°C (b) 573 – 273 = 300°C [/solution]...
Read More →An isosceles triangle has perimeter 30 cm
[question] Question. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle. [/question] [solution] Solution: Let the third side of this triangle be x. Perimeter of triangle = 30 cm $12 \mathrm{~cm}+12 \mathrm{~cm}+x=30 \mathrm{~cm}$ $x=6 \mathrm{~cm}$ $s=\frac{\text { Perimeter of triangle }}{2}=\frac{30 \mathrm{~cm}}{2}=15 \mathrm{~cm}$ By Heron’s formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=[\sqrt{15(15-12)(15-12)(15-6)}] \mathr...
Read More →Sides of a triangle are in the ratio of 12: 17: 25
[question] Question. Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area. [/question] [solution] Solution: Let the common ratio between the sides of the given triangle be x. Therefore, the side of the triangle will be $12 x, 17 x$, and $25 x$. Perimeter of this triangle $=540 \mathrm{~cm}$ $12 x+17 x+25 x=540 \mathrm{~cm}$ $54 x=540 \mathrm{~cm}$ $x=10 \mathrm{~cm}$ Sides of the triangle will be $120 \mathrm{~cm}, 170 \mathrm{~cm}$, and $250 \mathrm{~cm}...
Read More →Find the area of a triangle two sides of which are 18 cm
[question] Question. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm. [/question] [solution] Solution: Let the third side of the triangle be x. Perimeter of the given triangle = 42 cm $18 \mathrm{~cm}+10 \mathrm{~cm}+x=42$ $x=14 \mathrm{~cm}$ $s=\frac{\text { Perimeter }}{2}=\frac{42 \mathrm{~cm}}{2}=21 \mathrm{~cm}$ By Heron's formula, Area of a triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Area of the given triangle $=(\sqrt{21(21-18)(21-10)(21-14)}) \mathrm...
Read More →There is a slide in the park.
[question] Question. There is a slide in the park. One of its side walls has been painted in the same colour with a message “KEEP THE PARK GREEN AND CLEAN” (see the given figure). If the sides of the wall are 15m, 11m, and 6m, find the area painted in colour. [/question] [solution] Solution: t can be observed that the area to be painted in colour is a triangle, having its sides as 11 m, 6 m, and 15 m. Perimeter of such a triangle $=(11+6+15) \mathrm{m}$ $2 s=32 \mathrm{~m}$ $s=16 \mathrm{~m}$ By...
Read More →A traffic signal board, indicating ‘SCHOOL AHEAD’
[question] Question. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board? [/question] [solution] Solution: Side of traffic signal board = a Perimeter of traffic signal board $=3 \times a$ $2 s=3 a \Rightarrow s=\frac{3}{2} a$ By Heron’s formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Area of given triangle $=\sqrt{\frac{3}{2...
Read More →Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why.
[question] Question Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why. [/question] [solution] Solution Ice is a solid but its density is lower than water due to its structure. The molecules in ice make a cage like structure with lot of vacant spaces, this makes ice float on water. [/solution]...
Read More →Construct a right triangle whose base is 12 cm
[question] Question. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. [/question] [solution] Solution: The below given steps will be followed to construct the required triangle. Step I: Draw line segment AB of 12 cm. Draw a ray AX making 90° with AB. Step II: Cut a line segment AD of 18 cm (as the sum of the other two sides is 18) from ray AX. Step III: Join DB and make an angle DBY equal to ADB. Step IV: Let BY intersect AX at C. Join AC, BC. ΔAB...
Read More →Construct a triangle XYZ in which ∠Y = 30°,
[question] Question. Construct a triangle $X Y Z$ in which $\angle Y=30^{\circ}, \angle Z=90^{\circ}$ and $X Y+Y Z+Z X=11 \mathrm{~cm}$. [/question] [solution] Solution: The below given steps will be followed to construct the required triangle. Step I: Draw a line segment AB of 11 cm. (As $X Y+Y Z+Z X=11 \mathrm{~cm}$ ) Step II: Construct an angle, ∠PAB, of 30° at point A and an angle, ∠QBA, of 90° at point B. Step III: Bisect ∠PAB and ∠QBA. Let these bisectors intersect each other at point X. S...
Read More →Construct a triangle PQR in which QR = 6 cm,
[question] Question. Construct a triangle $P Q R$ in which $Q R=6 \mathrm{~cm}, \angle Q=60^{\circ}$ and $P R-P Q=2 \mathrm{~cm}$ [solution] Solution: The below given steps will be followed to construct the required triangle. Step I: Draw line segment $Q R$ of $6 \mathrm{~cm}$. At point $Q$, draw an angle of $60^{\circ}$, say $\angle X Q R$. Step II: Cut a line segment $Q S$ of $2 \mathrm{~cm}$ from the line segment $Q T$ extended in the opposite side of line segment $X Q$. (As PR PQ and PR $P Q...
Read More →Give reasons :
[question] Give reasons : A gas fills completely the vessel in which it is kept. A gas exerts pressure on the walls of the container. A wooden table should be called a solid. We can easily move our hand in air but to do the same through a solid block of wood we need a karate expert. [/question] [solution] Solution The molecules of gas have high kinetic energy due to which they keep moving in all directions and hence fill the vessel completely in which they are kept. A gas exerts pressure on the ...
Read More →Construct a triangle ABC in which BC = 8 cm,
[question] Question. Construct a triangle $A B C$ in which $B C=8 \mathrm{~cm}, \angle B=45^{\circ}$ and $A B-A C=3.5 \mathrm{~cm}$. [/question] [solution] Solution: The below given steps will be followed to draw the required triangle. Step I: Draw the line segment $B C=8 \mathrm{~cm}$ and at point $B$, make an angle of $45^{\circ}$, say $\angle X B C$. Step II: Cut the line segment $B D=3.5 \mathrm{~cm}$ (equal to $A B-A C$ ) on ray $B X$. Step III: Join DC and draw the perpendicular bisector P...
Read More →Construct an equilateral triangle,
[question] Question. Construct an equilateral triangle, given its side and justify the construction [/question] [solution] Solution: Let us draw an equilateral triangle of side 5 cm. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be 5 cm. We also know that each angle of an equilateral triangle is 60º. The below given steps will be followed to draw an equilateral triangle of 5 cm side. Step I: Draw a line segment AB of 5 cm lengt...
Read More →Construct the following angles and verify by measuring them by a protractor:
[question] Question. Construct the following angles and verify by measuring them by a protractor: (i) $75^{\circ}$ (ii) $105^{\circ}$ (iii) $135^{\circ}$ [/question] [solution] Solution: (i) $75^{\circ}$ The below given steps will be followed to construct an angle of 75°. (1) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R. (2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S. (3...
Read More →Construct the angles of the following measurements:
[question] Question. Construct the angles of the following measurements: (i) $30^{\circ}$ (ii) $22 \frac{1}{2}$ 。 (iii) $15^{\circ}$ [/question] [solution] Solution: (i) $30^{\circ}$ The below given steps will be followed to construct an angle of 30°. Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc of a circle which intersects PQ at R. Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S. ...
Read More →construct an angle of 45° at the initial point of a given
[question] Question. Construct an angle of 45° at the initial point of a given ray and justify the construction. [/question] [solution] Solution: The below given steps will be followed to construct an angle of 45°. (i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R. (ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S. (iii) Taking S as centre and with the same radius as before...
Read More →Construct an angle of 90° at the initial point of a given ray and justify the construction.
[question] Question. Construct an angle of 90° at the initial point of a given ray and justify the construction. [/question] [solution] Solution: The below given steps will be followed to construct an angle of 90°. (i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R. (ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S. (iii) Taking S as centre and with the same radius as before...
Read More →In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect,
[question] Question. In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC. [/question] [solution] Solution: Let perpendicular bisector of side BC and angle bisector of ∠A meet at point D. Let the perpendicular bisector of side BC intersect it at E. Perpendicular bisector of side BC will pass through circumcentre O of the circle. ∠BOC and ∠BAC are the angles subtended by arc BC at the centre...
Read More →Two congruent circles intersect each other at points A and B.
[question] Question. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ. [/question] [solution] Solution: $\mathrm{AB}$ is the common chord in both the congruent circles. $\therefore \angle \mathrm{APB}=\angle \mathrm{AQB}$ In $\triangle B P Q$, $\angle \mathrm{APB}=\angle \mathrm{AQB}$ $\therefore B Q=B P$ (Angles opposite to equal sides of a triangle) [solution]...
Read More →AC and BD are chords of a circle which bisect each other. Prove that
[question] Question. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle. [/question] [solution] Solution: Let two chords AB and CD are intersecting each other at point O. In $\triangle \mathrm{AOB}$ and $\triangle \mathrm{COD}$,In $\triangle \mathrm{AOB}$ and $\triangle \mathrm{COD}$, $O A=O C$ (Given) $O B=O D$ (Given) $\angle A O B=\angle C O D$ (Vertically opposite angles) $\triangle \mathrm{AOB} \cong \triangle \mathrm{C...
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