Geometry

We know that the point of concurrency of the medians of a triangle is known as centroid which is denoted as G.

We know that the point of concurrency of the altitudes of a triangle is known as orthocentre.

We know that the point of concurrency of the angle bisectors of a triangle is known as incentre.

We know that the point of concurrency of the perpendicular bisectors of a triangle is known as circumcentre.

In an isosceles triangle, AB = AC and ∠B = 65°.

We know that angles opposite to equal sides are equal.

∴ ∠B = ∠C = 65°

We know that the sum of all angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 65° + 65° = 180°

⇒ ∠A = 180° - 130°

∴ ∠A = 50°

We know that the side opposite to the shortest angle is the shortest.

∴ BC is the shortest side.

Given in a right-angled ΔPQR, ∠P = 90°, PQ = 10 cm and PR = 24 cm.

We know that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

⇒ In ΔPQR, QR² = PQ² + PR²

⇒ QR² = 10² + 24²

= 100 + 576

= 676

∴ QR = √676 = 26 cm

In a right angled ΔABC, AB = 25 cm, BC = 24 cm and AC = 7 cm.

We know that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

⇒ In ΔABC, AB² = BC² + AC²

⇒ 25² = 24² + 7²

⇒ 625 = 576 + 49

⇒ 625 = 625

∴ The given sides can be the sides of a right angled triangle.

In ΔPQR, ∠Q = 25° and ∠R = 65°.

We know that sum of angles of a triangle = 180°.

⇒ ∠P + ∠Q + ∠R = 180°

⇒ ∠P + 25° + 65° = 180°

⇒ ∠P = 180° - 90°

∴ ∠P = 90°

∴ Δ PQR is a right angled triangle which is right angled at P.

Given, PQ = 4 cm and PR = 3 cm

We know that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

⇒ In ΔPQR, QR² = PQ² + PR²

⇒ QR² = 4² + 3²

= 16 + 9

= 25

∴ QR = √25 = 5 cm

Let BC be the length of ladder i.e. 15 cm and the height of the window from the ground be AB i.e. 12 cm.

We know that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

⇒ In ΔABC, BC² = AB² + AC²

⇒ 15² = 12² + x²

⇒ 225 = 144 + x²

⇒ x² = 225 – 144 = 81

∴ x = √81 = 9 cm

Let us consider ΔABC an equilateral triangle with side 10 cm.

Construction: Draw a perpendicular bisector at C to AB such that AD = DB = 5 cm and the triangle is cut into two halves.

We have to find the altitude i.e. CD.

Consider ΔBCD,

We know that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

⇒ BC² = BD² + CD²

⇒ 10² = 5² + CD²

⇒ 100 = 25 + CD²

⇒ CD² = 100 – 25 = 75

∴ CD = √75 = = 5√3 cm

∴ Altitude = 5√3 cm

We know that the numbers which are satisfying the Pythagoras theorem are called the Pythagorean triplets.

We know that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In ΔPQR,

⇒ QR² = PQ² + PR²

⇒ 13² = 12² + 5²

⇒ 169 = 144 + 25

∴ 169 = 169

Since the above numbers satisfy the Pythagoras Theorem, 15, 5 and 13 form a Pythagorean Triplet.

We know that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Consider ΔABC,

⇒ BC² = AB² + AC²

⇒ BC² = 16² + 12²

⇒ BC² = 256 + 144

⇒ BC² = 400

∴ BC = √400 = 20 cm (Length of ladder)

Now after the ladder being pushed 2 feet farther, consider ΔBDE,

⇒ DE² = BD² + BE²

⇒ 20² = 14² + BE²

⇒ BE² = 400 – 196 = 204

⇒ BE = √204 = = 2√51 cm

∴ The ladder reaches 2√51 cm far up side the house.

We know that the constant distance from the centre of the circle is known as the radius.

In the figure, OA = radius.

In the figure, POQ is a diameter of the circle.

O is the midpoint of PQ and OP = OQ = radius of circle

∴ Diameter = 2 × Radius

A diameter is a chord that passes through the center of the circle.

Hence, the diameter is the longest chord of a circle.

In a circle, the sum of two diameters = 200 mm

Let diameter be d and radius be r of a circle.

⇒ d + d = 200 mm

⇒ 2d = 200 mm

∴ d = 100 mm

We know that Diameter = 2(Radius).

⇒ 100 = 2r

⇒ r = 100 ÷ 2

∴ r = 50 mm

We know that 1 cm = 10 mm.

⇒ r = 50 ÷ 10 = 5 cm

∴ The radius of the circle is 5 cm.

Circle Segment: A chord of a circle divides the circular region into two parts. Each part is called as segment of a circle.

The segment containing minor arc is called the minor segment.

The segment containing major arc is called the major segment.

Sector of a Circle:

The circular region enclosed by an arc of a circle and the two radii at its end points is known as sector of a circle.

The smaller sector OALB is called the minor sector.

The greater sector OAMB is called the major sector.

Here, in the figure, AB is the chord. The chord AB divides the circle into two parts.

The curved parts ALB and AMB are known as arcs.

The smaller arc ALB is the minor arc.

The greater arc AMB is the major arc.

Tangent:

Tangent is a line that touches a circle at exactly one point, and the point is known as point of contact.

In the figure, EF is a tangent and G is the point of contact.

Secant:

A line passing through a circle and intersecting the circle at two points is called the secant of a circle.

In the figure AB and CD are secants.