Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 18 Constructions

Exercise 18.1

Answer 1.
Answer:
Steps of constructions:
(i) Draw a circle of radius 4 cm with centre O.
(ii) Join the centre O to the given point P.
(iii) On the given point P, draw a perpendicular to OP.
(iv) PT is the required tangent.
In simple words: We draw a circle, connect its center to the outside point, then draw a line at right angles to this connection. This gives us a tangent line that touches the circle.

[Diagram: This diagram shows a circle with center O and radius 4 cm, with point P outside the circle. A line OP connects the center to point P, and a perpendicular line PT is drawn from P to create a tangent to the circle.]

📝 Teacher's Note: Show students that a tangent always makes a 90-degree angle with the radius at the point of contact. Use a string and coin to demonstrate this in class.

🎯 Exam Tip: Always write "perpendicular to OP" clearly. The examiner wants to see you know that tangent and radius are at right angles.

Answer 2.
Answer:
Steps of construction:
(i) Draw a circle with radius 4.5 cm.
(ii) At any point P draw a chord PA.
(iii) Take any point B on the circle and join PB and AB.
(iv) At P, draw ∠APT equal to ∠ABP.
(v) PT is the required tangent.
In simple words: We draw a circle and a chord. Then we make an angle at P that matches an angle made by another point on the circle. This creates the tangent line.

[Diagram: This diagram shows a circle with a chord PA, point B on the circle, and angle APT constructed equal to angle ABP to form tangent PT.]

📝 Teacher's Note: Explain that this method uses the alternate segment theorem. The angle between tangent and chord equals the angle in the alternate segment.

🎯 Exam Tip: Write "∠APT = ∠ABP" clearly. This equal angle relationship is the key step that creates the tangent.

Answer 3.
Answer:
Steps of construction:
(i) Draw a circle of radius 2.5 cm with centre O.
(ii) Join the centre O to the given point P which is 6 cm away from O.
(iii) Draw a perpendicular bisector of OP. Let M be the mid-point of OP.
(iv) With M as centre and radius OM, draw a circle cutting the first circle at A and B.
(v) Join PA and PB.
(vi) PA and PB are the required tangents.
In simple words: We connect the outside point to the center, find the middle point, then draw a helper circle. Where this helper circle cuts the main circle gives us the tangent points.

[Diagram: This diagram shows two circles - one with center O and radius 2.5 cm, and another with center M (midpoint of OP) that intersects the first circle at points A and B, creating tangents PA and PB.]

📝 Teacher's Note: Tell students that this method always gives two tangents from any outside point. Show them how the helper circle acts like a guide.

🎯 Exam Tip: Write "perpendicular bisector of OP" and "M is mid-point of OP" clearly. These are the key construction steps that get marks.

Answer 4.
Answer:
Steps of construction:
(i) Draw a circle of radius 3 cm with centre O.
(ii) If P is the given point, then draw PAB a secant to the given circle.
(iii) Draw a perpendicular bisector of PB and let M be the mid-point of PB.
(iv) With M as centre and MP as radius, draw a semi-circle on PB.
(v) At A, draw a perpendicular to PB. Let this perpendicular meet the semi-circle at D.
(vi) With P as centre and PD as radius, cut off two arcs on the given circle at T and S.
(vii) Join PT and PS.
(viii) PT and PS are the required tangents.
In simple words: We draw a line through the circle, make a helper semi-circle, find a special length, then use this length to find where the tangents touch the circle.

[Diagram: This diagram shows a circle with center O, a secant PAB, a semi-circle construction with center M, and the resulting tangents PT and PS.]

📝 Teacher's Note: This is a more complex construction. Explain each step slowly and show how the semi-circle helps us find the correct tangent length.

🎯 Exam Tip: Draw the perpendicular bisector and semi-circle very clearly. These helper constructions are crucial for getting the right answer.

Answer 5.
Answer:
Steps of construction:
(i) Draw a circle of radius 3 cm with centre O.
(ii) Join the centre O to the given point P which is 5 cm away from O.
(iii) Draw a perpendicular bisector of OP. Let M be the mid-point of OP.
(iv) With M as centre and radius OM, draw a circle cutting the first circle at A and B.
(v) Join PA and PB.
(vi) PA and PB are the required tangents.
(vii) On measuring, PA and PB = 4 cm
In simple words: This is the same method as Answer 3, but with different measurements. We find that both tangent lines are exactly 4 cm long.

[Diagram: This diagram shows a circle with center O and radius 3 cm, point P at 5 cm distance, and the construction showing tangents PA and PB, each measuring 4 cm.]

📝 Teacher's Note: Point out that both tangents from an external point are always equal in length. This is an important property students should remember.

🎯 Exam Tip: Always measure and write the length of tangents when asked. Write "PA = PB = 4 cm" to show you know both tangents are equal.

Answer 6.
Answer:
Steps of construction:
(i) Draw a line OP = 8 cm.
(ii) At O, draw a circle of radius 3.5 cm.
(iii) At P, draw a circle of radius 2.5 cm.
(iv) At O, draw a third circle concentric to the bigger circle and radius = (3.5 - 2.5) cm = 1 cm
(v) Draw a perpendicular bisector of OP. Let R be the mid-point of OP.
(vi) With R as centre and OR as radii, draw a fourth circle. Mark as T and S where the third and fourth circles intersect each other.
(vii) Join OT and OS and extend lines to meet the bigger circle at A and B.
(viii) Join PT and PS.
(ix) On PT and PS, draw perpendiculars to meet the smaller circle at M and N.
(x) Join AM and BN.
AM and BN are the required tangents.
In simple words: This constructs common tangents to two circles. We use helper circles and perpendicular bisectors to find the exact points where the tangent lines should touch both circles.

[Diagram: This diagram shows two circles of different sizes with centers O and P, and the construction of common external tangents AM and BN using multiple helper circles.]

📝 Teacher's Note: This is the most complex construction. Break it into smaller steps and show students how each helper circle serves a purpose in finding the tangent points.

🎯 Exam Tip: Draw all construction lines lightly first, then darken the final tangent lines AM and BN. Label all points clearly to avoid confusion.

Answer 7.
Answer:
Steps of construction:
(i) Draw a line OP = 6 cm.
(ii) At O, draw a circle of radius 3.5 cm.
(iii) At P, draw a circle of radius 2 cm.
(iv) At O, draw a third circle concentric to the bigger circle and radius = (3.5 - 2) cm = 1.5 cm
(v) Draw a perpendicular bisector of OP. Let R be the mid-point of OP.
(vi) With R as centre and OR as radii, draw a fourth circle. Mark as T and S where the third and fourth circles intersect each other.
(vii) Join OT and OS and extend lines to meet the bigger circle at A and B.
(viii) Join PT and PS.
(ix) On PT and PS, draw perpendiculars to meet the smaller circle at M and N.
(x) Join AM and BN.
AM and BN are the required tangents.

Proof:
Since AT || PM and BS || PN; therefore AM = PT and BN = PS
Now in △OTP and △OSP
OT = OS (Tangents to a circle from same point)
Therefore, AM = BN
Hence, proved.
In simple words: This is similar to Answer 6 but with different measurements. The proof shows that the two common tangents we construct are equal in length because of the properties of tangent lines.

[Diagram: This diagram shows two circles with centers O and P, and the complete construction of common external tangents AM and BN with all the helper construction lines.]

📝 Teacher's Note: The proof uses parallel lines and tangent properties. Explain that tangents from the same external point are always equal, which makes AM = BN.

🎯 Exam Tip: When asked to prove, write "OT = OS (Tangents from same point)" clearly. This is the key property that makes the proof work.

Answer 8.
Answer:

[Diagram: This diagram shows a geometric construction with multiple intersecting circles and lines, demonstrating the steps to construct tangents from an external point to a circle.]

Steps of construction:
(i) Draw a line OP = 8 cm.
(ii) At O, draw a circle of radius 3 cm.
(iii) At P, draw a circle of radius 3.5 cm.
(iv) At O, draw a third circle concentric to the smaller circle and radius = (3.5 + 3) cm = 6.5 cm
(v) Draw a perpendicular bisector of OP. Let R be the mid-point of OP.
(vi) With R as centre and OR as radii, draw a fourth circle. Mark as T and S where the third and fourth circles intersect each other.
(vii) Join OT and OS to meet the smaller circle at A and B.
(viii) Join PT and PS.
(ix) On PT and PS, draw perpendiculars to meet the bigger circle at M and N.
(x) Join AM and BN.
AM and BN are the required tangents.
On measuring, AM = BN = 8 cm.
In simple words: We draw circles and find special points where lines touch the circles. These touching lines are called tangents. We use geometry steps to find the exact places.

📝 Teacher's Note: Show students how a tangent touches a circle at exactly one point. Use a coin and a ruler to demonstrate this in class.

🎯 Exam Tip: Always draw neat diagrams and label all points clearly. Write the final answer separately at the end.

Answer 9.
Answer:

[Diagram: This diagram shows another geometric construction with intersecting circles demonstrating the construction of transverse common tangents.]

Steps of construction of transverse common tangent:
(i) Draw a line OP = 7 cm.
(ii) At O, draw a circle of radius 2.5 cm.
(iii) At P, draw a circle of radius 4 cm.
(iv) At O, draw a third circle concentric to the smaller circle and radius = (2.5 + 4) cm = 6.5 cm
(v) Draw a perpendicular bisector of OP. Let R be the mid-point of OP.
(vi) With R as centre and OR as radii, draw a fourth circle. Mark as T and S where the third and fourth circles intersect each other.
(vii) Join OT and OS to meet the smaller circle at A and B.
(viii) Join PT and PS.
(ix) On PT and PS, draw perpendiculars to meet the bigger circle at M and N.
(x) Join AM and BN.
AM and BN are the required tangents.
In simple words: Transverse tangents cross between two circles. We follow special steps to find where these crossing lines touch each circle perfectly.

📝 Teacher's Note: Explain that transverse tangents cross between circles while direct tangents do not cross. Draw both types to show the difference.

🎯 Exam Tip: Remember transverse tangents cross between circles. Write this key point and draw the crossing clearly in your diagram.

Answer 10.
Answer:

[Diagram: This diagram shows a triangle ABC inscribed in a circle, with construction lines showing the circumcircle construction process.]

Steps of construction:
(i) Draw line AB = 5 cm
(ii) At B, draw an arc making an angle of 60° with AB
(iii) On the arc cut BC = 4.5 cm.
(iv) Join AC.
(v) Draw perpendicular bisectors of AB and BC, which meet at O.
(vi) With O as centre and radius equal to the distance between O and the vertex of the triangle, draw a circle to pass through all the three vertices of the triangle.
(vii) The drawn circle is the required circle with radius = 2.8 cm
In simple words: We make a triangle first. Then we find the center point that is equal distance from all three corners. We draw a circle from this center that passes through all three corners.

📝 Teacher's Note: Show students that the circumcenter is where perpendicular bisectors meet. This point is equal distance from all vertices.

🎯 Exam Tip: Always find the circumcenter by drawing perpendicular bisectors. Measure the radius carefully and write the final measurement.

Answer 11.
Answer:

[Diagram: This diagram shows a triangle with its circumcircle, demonstrating the construction of a circumscribed circle around a triangle.]

Steps of construction:
(i) Draw line AB = 5 cm
(ii) At B, draw an arc with radius 6 cm
(iii) On the arc cut AC = 4.5 cm.
(iv) Join AC and BC.
(v) Draw perpendicular bisectors of AB and BC, which meet at O.
(vi) With O as centre and radius equal to the distance between O and the vertex of the triangle, draw a circle to pass through all the three vertices of the triangle.
(vii) The circle drawn is the required circle.
In simple words: We build a triangle step by step. Then we find the special center point and draw a circle that goes through all three corners of the triangle.

📝 Teacher's Note: Use a compass to show how the circumcenter is equal distance from all three vertices. This makes the circle possible.

🎯 Exam Tip: Draw perpendicular bisectors neatly and mark the circumcenter clearly. The final circle must pass through all three vertices exactly.

Answer 12.
Answer:

[Diagram: This diagram shows an equilateral triangle inscribed in a circle, with a 60° angle marked and construction lines visible.]

Steps of construction:
(i) Draw line AB = 4.5 cm
(ii) At B, draw an arc making an angle of 60° with AB and length BC = 4.5 cm.
(iii) On the arc cut AC = 4.5 cm.
(iv) Join AC.
(v) Draw perpendicular bisectors of AB and BC, which meet at O.
(vi) With O as centre and radius equal to the distance between O and the vertex of the triangle, draw a circle to pass through all the three vertices of the triangle.
(vii) The drawn circle is the required circle with radius = 2.6 cm
In simple words: We make a triangle where all sides are equal (equilateral triangle). Then we draw a circle that touches all three corners of this triangle.

📝 Teacher's Note: Explain that when all sides are equal and the angle is 60°, we get an equilateral triangle. The circumcircle is very neat for such triangles.

🎯 Exam Tip: For equilateral triangles, all angles are 60° and all sides are equal. Write the final radius measurement clearly with units.

Answer 13.
Answer:
Steps of construction:
i) Draw a line segment BC = 7.5 cm
ii) At B, draw an arc making an angle of 60° with BC
iii) At C, draw an arc with radius (AC = AB + 1.5 cm) = 9 cm cutting the previous arc.
iv) Join AC and AB.
v) Draw angle bisectors for ∠A and ∠B meeting at O.
vi) Draw a perpendicular to BC from O and mark it as M.
vii) With OM as radius draw a circle touching all three sides of the triangle.
viii) The drawn circle is the required circle with radius = 2.3 cm

[Diagram: This diagram shows triangle ABC with an inscribed circle. The circle touches all three sides of the triangle at points, and O is the center of the inscribed circle.]

In simple words: We made a triangle first. Then we found the center by drawing angle bisectors. From that center, we drew a circle that touches all three sides inside the triangle.

📝 Teacher's Note: Show students that the inscribed circle always touches all three sides. The center is where angle bisectors meet. This point is always the same distance from all three sides.

🎯 Exam Tip: Always draw angle bisectors to find the center. Mark the touching points clearly. Write "inscribed circle" in your answer to get full marks.

Answer 14.
Answer:
Steps of construction:
(i) Draw a circle with centre O and radius = 3 cm.
(ii) Draw radii OA and OB such that ∠AOB = (360/3) = 120°
(iii) Join AB. Cut off arcs AC and BC equal to AB.
(iv) Join AC and BC.
△ABC is the required equilateral triangle.

[Diagram: This diagram shows a circle with center O, and an equilateral triangle ABC inscribed in it, where all three vertices A, B, C lie on the circle.]

In simple words: We drew a circle first. Then we marked three points on the circle at equal distances. When we joined these points, we got a triangle with all sides equal.

📝 Teacher's Note: Explain that 360° ÷ 3 = 120° because we need three equal parts around the circle. Each angle at the center is 120°.

🎯 Exam Tip: Write the angle calculation clearly: 360°/3 = 120°. Always mention that the triangle is equilateral in your final answer.

Answer 15.
Answer:
Steps of construction:
(i) Draw a circle of radius 2.5 cm with centre O.
(ii) Draw two diameters PQ and RS of the circle meeting at centre O.
(iii) Taking OP as radius cut two arcs from P on both sides (left and right) of P. Repeat same with Q, R and S.
(iv) Mark new points formed as A, B, C and D.
(v) Join AB, BC, CD and AD.
ABCD is the required square circumscribing the given circle.

[Diagram: This diagram shows a circle with center O, two perpendicular diameters PQ and RS, and a square ABCD drawn outside the circle such that all four sides of the square touch the circle.]

In simple words: We drew a circle and two lines through the center. Then we drew a square around the circle so that all four sides of the square just touch the circle.

📝 Teacher's Note: Show students that circumscribing means "drawing around." The square is outside the circle and touches it at four points.

🎯 Exam Tip: Write "circumscribing" correctly. Make sure your square touches the circle at exactly four points - one on each side.

Answer 16.
Answer:
Steps of construction:
(i) Draw a circle with centre O and radius = 3.5 cm.
(ii) Draw radii OA and OB such that ∠AOB = (360/3) = 120°
(iii) Cut off arcs BC, CD, DE, EF and AF equal to AB.
(iv) Join AB, BC, CD, DE, EF and AF.
ABCDEF is the required regular hexagon inscribed in the given circle.

[Diagram: This diagram shows a circle with center O and a regular hexagon ABCDEF inscribed in it, where all six vertices lie on the circle.]

In simple words: We drew a circle and marked six equal points on it. When we joined these points in order, we got a shape with six equal sides - that is a hexagon.

📝 Teacher's Note: Explain that 360° ÷ 6 = 60° for a hexagon. Each central angle is 60°. A hexagon has 6 sides, all equal in a regular hexagon.

🎯 Exam Tip: Write "regular hexagon" clearly. Show the calculation 360°/6 = 60° for the central angle. All sides must be equal.

Answer 17.
Answer:
Steps of construction:
(i) Draw a circle with centre O and radius = 3 cm.
(ii) Draw radii OA and OB such that ∠AOB = (360/5) = 72°
(iii) Cut off arcs BC, CD, DE and AE equal to AB.
(iv) Draw tangents to the circle at A, B, C, D and E.
(v) Let these tangents intersect at P, Q, R, S and T.
PQRST is the required regular pentagon.

[Diagram: This diagram shows a circle with center O, five points A, B, C, D, E on the circle, and a regular pentagon PQRST formed by tangents to the circle at these points.]

In simple words: We marked five equal points on a circle. Then we drew lines that just touch the circle at these points. These lines met to form a pentagon outside the circle.

📝 Teacher's Note: A pentagon has 5 sides. Show students that 360° ÷ 5 = 72°. The tangent lines meet to form the pentagon outside the circle.

🎯 Exam Tip: Write the angle calculation 360°/5 = 72°. Draw tangents carefully - they should just touch the circle at one point each.

Answer 18.
Answer:
Steps of construction:
(i) Draw a circle with centre O and radius = 4 cm.
(ii) Draw radii OT and OP such that ∠TOP = (360/5) = 72°
(iii) Cut off arcs PQ, QR, RS, ST equal to TP.
(iv) Join TP, PQ, QR, RS and ST.
PQRST is the required regular hexagon inscribed in the given circle.
(v) From centre O, drop perpendiculars on TP, PQ, QR, RS and ST at A, B, C, D and E.
(vi) With OA as radius draw a circle touching the five sides of the pentagon.
The circle drawn is the required circle with radius = 3.3 cm

[Diagram: This diagram shows a circle with center O, a regular pentagon PQRST inscribed in it, and a smaller inscribed circle inside the pentagon that touches all five sides.]

In simple words: First we made a pentagon inside a circle. Then we drew a smaller circle inside the pentagon that touches all five sides of the pentagon.

📝 Teacher's Note: This shows both circumscribed (big circle around pentagon) and inscribed (small circle inside pentagon) circles. The pentagon is between both circles.

🎯 Exam Tip: Draw both circles clearly. Label which circle is circumscribed and which is inscribed. Write the final radius value.

Answer 19.
Answer:
Steps of construction:
(i) A 30 degree angle can be formed by knowing that the inverse sine of 0.5 is 30 degrees. In other words, a right triangle with a 30 degree angle has the hypotenuse twice as long as the leg opposite the near leg.
(ii) Using your compass, construct a 90°, then construct the leg opposite the 30 degree angle. Construct the hypotenuse twice as long, that makes a 30 degree angle.
(iii) Bisect 30 degree angle and you have 15 degrees.
(iv) From 90°, cut off angle equal to 15° angle to get 75° angle. Bisect the 75° angle to get 37.5° angle.
(v) Now draw angle bisector of angle 37.5°.
(vi) On OR, at 5 cm from O, drop a perpendicular PA from the angle bisector of angle 37.5°.
(vii) With PA as radius, draw a circle touching OA at A and OB at B.
This is the required circle.

[Diagram: This diagram shows angle construction with various angle measurements and a circle touching two sides of an angle.]

In simple words: We made specific angles step by step using a compass. Then we found a special point and drew a circle from there that touches two lines of our angle.

📝 Teacher's Note: Start with simple angles like 90° and 60°. Show students how to bisect angles to get smaller angles. Practice angle construction before attempting this problem.

🎯 Exam Tip: Draw all construction lines clearly. Mark all angles with their measurements. Show the circle touching both lines of the angle.

Answer 20.

[Diagram: This diagram shows a geometric construction with a circle, points labeled O, A, B, M, and N, with various lines and arcs drawn to construct a circle touching two given points.]

Answer:

Steps of construction:

1. Draw a OA = 4.5 cm
2. At O, draw an arc with angle 45 degree and radius 3.2 cm
3. Join AB
4. At A, draw a perpendicular to OA
5. Bisect AB and let bisector meet the perpendicular to OA at M
6. with MA as radius draw a circle touching B and A.

The circle drawn is the required circle.

In simple words: This is a step-by-step method to draw a circle that touches two specific points. We use perpendicular lines and bisectors to find the exact center of the circle. Then we draw the circle with the right radius.

📝 Teacher's Note: Show students how to use a compass and ruler for each step. The key is finding the center point M first. Practice drawing perpendicular lines and bisectors before this construction.

🎯 Exam Tip: Write each step clearly with proper labels. Draw neat diagrams with all points marked. Always end with "The circle drawn is the required circle" to show your construction is complete.