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Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (ΔPAB and ΔPCD) and the green region (ΔPBC and ΔPDA)?
Let the side length of square ABCD be s. Draw a line through point P perpendicular to AB and CD. Let the perpendicular distance from P to AB be h1 and to CD be h2. Then h1 + h2 = s. Area of red region = Area(ΔPAB) + Area(ΔPCD) = (1/2) x s x h1 + (1/2) x s x h2 = (1/2) x s x (h1 + h2) = (1/2) x s². SRead more
Let the side length of square ABCD be s.
Draw a line through point P perpendicular to AB and CD.
Let the perpendicular distance from P to AB be h1 and to CD be h2.
Then h1 + h2 = s.
Area of red region = Area(ΔPAB) + Area(ΔPCD)
= (1/2) x s x h1 + (1/2) x s x h2 = (1/2) x s x (h1 + h2) = (1/2) x s².
Since total area of the square is s², the green region also equals s² – s²/2 = (1/2) x s².
Thus, the ratio of areas is 1:1.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessIn ΔABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area (ΔABP) = area (ΔACP).
Since D is the midpoint of BC, AD is a median of triangle ABC. A median divides a triangle into two triangles of equal area: Area(ΔABD) = Area(ΔACD) ... (Equation 1). In triangle PBC, PD is also a median because D is the midpoint of BC. Therefore: Area(ΔPBD) = Area(ΔPCD) ... (Equation 2). SubtractinRead more
Since D is the midpoint of BC, AD is a median of triangle ABC.
A median divides a triangle into two triangles of equal area:
Area(ΔABD) = Area(ΔACD) … (Equation 1).
In triangle PBC, PD is also a median because D is the midpoint of BC.
Therefore:
Area(ΔPBD) = Area(ΔPCD) … (Equation 2).
Subtracting Equation 2 from Equation 1:
Area(ΔABD) – Area(ΔPBD) = Area(ΔACD) – Area(ΔPCD)
Area(ΔABP) = Area(ΔACP).
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessIf the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)
Let ABCD be the 4-gon and E, F, G, H the midpoints of AB, BC, CD, DA respectively. Draw diagonal AC, dividing ABCD into triangles ABC and ADC. In triangle ABC, E and F are midpoints, so triangle EBF is similar to ABC with ratio 1/2, meaning Area(EBF) = (1/4) x Area(ABC). Similarly, Area(HDG) = (1/4)Read more
Let ABCD be the 4-gon and E, F, G, H the midpoints of AB, BC, CD, DA respectively.
Draw diagonal AC, dividing ABCD into triangles ABC and ADC.
In triangle ABC, E and F are midpoints, so triangle EBF is similar to ABC with ratio 1/2, meaning Area(EBF) = (1/4) x Area(ABC).
Similarly, Area(HDG) = (1/4) x Area(ADC).
Adding these:
Area(EBF) + Area(HDG) = (1/4) x [Area(ABC) + Area(ADC)] = (1/4) x Area(ABCD).
Similarly, using diagonal BD:
Area(HAE) + Area(GCF) = (1/4) x Area(ABCD).
Sum of four corner triangles = (1/4 + 1/4) x Area(ABCD) = (1/2) x Area(ABCD).
Subtracting corners gives:
Area(EFGH) = Area(ABCD) – (1/2) x Area(ABCD) = (1/2) x Area(ABCD).
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessA chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to πr²(1/6 – √3/4).
Note on textbook printing: The question contains a slight misprint where π is factored outside; the true difference is (1/6)πr² - (√3/4)r² = r²((π/6) - (√3/4)) = πr²(1/6 - √3/(4π)). Proof: Sector area with central angle 60°: Area(sector) = (60 / 360) x π x r² = (1/6) x π x r². The triangle formed byRead more
Note on textbook printing: The question contains a slight misprint where π is factored outside; the true difference is (1/6)πr² – (√3/4)r² = r²((π/6) – (√3/4)) = πr²(1/6 – √3/(4π)).
Proof:
Sector area with central angle 60°:
Area(sector) = (60 / 360) x π x r² = (1/6) x π x r².
The triangle formed by the chord and the two radii has two equal sides r and vertex angle 60°, so it is equilateral.
Area(triangle) = (√3 / 4) x r².
Area of minor segment = Area(sector) – Area(triangle)
= (1/6)πr² – (√3/4)r².
Writing with the textbook’s expression:
Area = πr²(1/6 – √3/4) (taking the intended algebraic form).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessWhen the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton’s third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig. 6.42). Explain why.
According to Newton's third law, the mutual magnetic forces are equal and opposite. However, the tiny compass needle has very small mass and is mounted on a near frictionless pivot, allowing the force to easily turn it. In contrast, the bar magnet has a much larger mass, higher inertia and experiencRead more
According to Newton’s third law, the mutual magnetic forces are equal and opposite. However, the tiny compass needle has very small mass and is mounted on a near frictionless pivot, allowing the force to easily turn it. In contrast, the bar magnet has a much larger mass, higher inertia and experiences significant friction from the resting tabletop, which prevents it from moving.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
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