This MCQ module is based on: Rectangles, Diagonals and Equidistant Points
TOPIC 31 OF 41
Rectangles, Diagonals and Equidistant Points
🎓 Class 6
Mathematics
CBSE
Theory
Ch 8 — Playing with Constructions
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Rectangles, Diagonals and Equidistant Points
Targeting Class 6 level in Geometry, with Basic difficulty.
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8.5 Rectangles with Diagonals at Specific Angles
The diagonal? of a rectangle divides the opposite corner angles (90°) into two parts. The two parts may be equal (45° + 45°) only if the rectangle is a square; otherwise they are unequal but always sum to 90°.
Observation
If a diagonal of a rectangle divides an opposite angle into \(x°\) and \((90-x)°\), then the ratio of adjacent sides depends on \(x\). When \(x=45°\), the sides are equal (square). When \(x=50°\) and \(90-x=40°\), the sides are in a specific non-equal ratio determined by \(\tan 50°\) and \(\tan 40°\) (to be learned later).
Construction (50° and 40°)
To build a rectangle in which the diagonal makes angles 50° and 40° with adjacent sides:
Rectangle ABCD with diagonal AC dividing ∠A into 50° + 40°.
Construct: Rectangle given One Side & the Diagonal
Construct a rectangle PQRS in which PQ = 4 cm and the diagonal PS = 8 cm.
Rectangle PQRS with side PQ = 4 cm, diagonal PS = 8 cm.
Procedure: Draw PQ = 4 cm. Erect perpendicular at Q. Now with compass centred at P and radius = 8 cm, draw an arc cutting the perpendicular at R. Finally erect perpendicular at P and complete the rectangle by marking S so that PS is parallel to QR. Check PS = 8 cm (this is actually the diagonal PR — we mark R and then find S by completing).
8.6 Points Equidistant from Two Given Points
When we look for all points equally distant from two fixed points A and B, we get a line — the perpendicular bisector? of AB. Every point on this line is the same distance from A as from B.
The locus of points equidistant from A and B is the perpendicular bisector of AB (dashed orange).
Construct: The "House" Figure (p.211)
Recreate a house shape with a rectangular base and a triangular roof, such that all lines forming the border of the house are of length 5 cm, and the apex of the roof A is equidistant from corners B and C (the top corners of the base).
House figure — all border sides 5 cm. Apex A is equidistant (5 cm) from B and C.
The first task is to locate A: it must be 5 cm from B and 5 cm from C simultaneously. Using a compass:
Step 1: With centre B and radius 5 cm, draw an arc above BC.
Step 2: With centre C and radius 5 cm, draw a second arc.
Step 3: The two arcs intersect at A.
In-text: Can any of the ideas used in the 'House' problem be used here (in a 4-sided figure where one point D should be 5 cm from B and C)?
Answer: Yes — use two compass arcs of radius 5 cm centred at B and C. The intersection point (below BC this time) is D.
Answer: Yes — use two compass arcs of radius 5 cm centred at B and C. The intersection point (below BC this time) is D.
Activity: Locus of Equidistant Points
L4 AnalyseMaterials: Ruler, compass, pencil, paper.
Predict: If you mark many points equidistant from two fixed dots A and B, what curve or line will they form?
- Mark A and B 8 cm apart.
- With radius 5 cm, draw arcs from A and B above and below AB. Mark the two intersections P (above) and Q (below).
- Repeat with radius 6 cm, radius 7 cm — get new pairs of points.
- Join all points you found. What do you notice?
All equidistant points lie on one straight line — the perpendicular bisector of AB. This line passes through the midpoint of AB and is at 90° to AB.
Equidistant Points Exercises
Q. Consider a 4-sided figure with corners B and C at the top and D and E at the bottom. Another vertex A is to be placed 5 cm from B and 5 cm from C. Where does A lie?
On the perpendicular bisector of BC, at distance 5 cm from each of B and C. Use two compass arcs of radius 5 cm from B and C; their intersection is A.
Competency-Based Questions
Scenario: A park has two old trees T1 and T2 exactly 10 m apart. A plan is made to place a fountain F that is equally distant from both trees, at a distance of 13 m from each.
Q1. Using a compass-and-ruler sketch, where must F be placed?
L3 ApplyF lies on the perpendicular bisector of T1T2, at the point 13 m from each tree. Draw arcs of radius 13 m centred at T1 and T2 — their intersection is F. (There are two such points, one on each side of the line T1T2.)
Q2. The gardener asks: "What is the perpendicular distance from F to the line T1T2?" Analyse using the right triangle formed.
L4 AnalyseThe midpoint M of T1T2 is 5 m from each tree. Triangle FMT1 is right-angled at M, with hypotenuse FT1 = 13 m and one leg MT1 = 5 m. So FM = \(\sqrt{13^2-5^2}=\sqrt{144}=12\) m.
Q3. Someone claims: "If I want F at 4 m from each tree, it's still possible." Evaluate.
L5 EvaluateImpossible. Two circles of radius 4 centred at T1 and T2 (10 m apart) cannot meet because 4 + 4 = 8 < 10. The minimum distance from each is 5 m (at the midpoint).
Q4. Design a triangular flower bed whose vertices are the two trees and a new fountain F that is 8 m from T1 and 6 m from T2. Create a brief plan.
L6 Create1. Draw T1T2 = 10 m. 2. Arc of radius 8 m centred at T1. 3. Arc of radius 6 m centred at T2. 4. Intersection = F. 5. Join T1F and T2F. Resulting triangle has sides 10, 8, 6 — a right triangle (since \(6^2+8^2=10^2\)), so the flower bed is right-angled at F.
Assertion–Reason Questions
A: The set of all points equidistant from two points A and B forms the perpendicular bisector of AB.
R: Every point on this bisector is the same distance from A as from B.
R: Every point on this bisector is the same distance from A as from B.
(a) — Both true, and R correctly states the defining property.
A: A diagonal of a rectangle divides it into two congruent right triangles.
R: Opposite sides of a rectangle are equal.
R: Opposite sides of a rectangle are equal.
(a) — Both true. Opposite sides being equal (with shared diagonal and right angles) gives two congruent right triangles by SSS/SAS.
A: Two circles of radius 4 cm whose centres are 10 cm apart intersect at two points.
R: Two circles intersect when the distance between centres is less than the sum of radii.
R: Two circles intersect when the distance between centres is less than the sum of radii.
(d) — A is false (4 + 4 = 8 < 10, so no intersection). R is true in general.
Frequently Asked Questions
Are the diagonals of a rectangle equal?
Yes. Both diagonals of any rectangle are exactly the same length, and they always intersect at the centre of the rectangle, dividing each other into two equal halves.
What does 'equidistant points' mean?
Points that are the same distance from a given reference point are equidistant. All points on a circle are equidistant from the centre, and all points on the perpendicular bisector are equidistant from the two ends of the segment.
How do you find points equidistant from two given points?
Draw the perpendicular bisector of the line segment joining the two points. Every point on that bisector is exactly the same distance from both original points.
Do square and rectangle diagonals behave the same way?
They share length equality and bisection at the centre, but the square adds one extra property: its diagonals are perpendicular and bisect the corners into 45-degree angles.
How is this topic used later?
Perpendicular bisectors and equidistance are the foundation for circumcircle construction, locus problems and coordinate geometry in higher classes.
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