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Practical Geometry - Construction of a Right-Angled Triangle (RHS criterion)

Grade 7CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The RHS criterion stands for Right angle-Hypotenuse-Side. To construct such a triangle, we need the length of one leg (base or height), the length of the hypotenuse, and the measure of the right angle (90∘90^{\circ}).

Components of an RHS triangle: Right angle, Hypotenuse, and one Side.
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In a right-angled triangle, the side opposite the 90∘90^{\circ} angle is the longest side, called the hypotenuse. According to the Pythagoras theorem, the square of the hypotenuse equals the sum of the squares of the other two sides: a2+b2=c2a^2 + b^2 = c^2.

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The construction process begins by drawing the base leg. Then, a perpendicular line (90∘90^{\circ}) is constructed at one endpoint using a protractor or compass. Finally, an arc of the hypotenuse's length is drawn from the other endpoint of the base to intersect the perpendicular line.

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The point of intersection between the hypotenuse arc and the perpendicular line defines the third vertex of the triangle.

Using a compass arc to find the vertex on the perpendicular line.

📐Formulae

Pythagoras Theorem: a2+b2=c2a^{2} + b^{2} = c^{2} (where cc is the hypotenuse)

Angle Sum Property: ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^{\circ}

In any right-angled triangle: Base2+Height2=Hypotenuse2\text{Base}^{2} + \text{Height}^{2} = \text{Hypotenuse}^{2}

💡Examples

Problem 1:

Construct a right-angled triangle ΔPQR\Delta PQR, right-angled at QQ, where QR=3 cmQR = 3\text{ cm} and hypotenuse PR=5 cmPR = 5\text{ cm}.

Solution:

  1. Draw a horizontal line segment QR=3 cmQR = 3\text{ cm} using a ruler.
  2. At point QQ, use a protractor or compass to draw a ray QXQX such that ∠RQX=90∘\angle RQX = 90^{\circ}. This ray should be perpendicular to QRQR.
  3. Set the compass to a radius of 5 cm5\text{ cm}. Place the compass pointer at point RR.
  4. Draw an arc that cuts the ray QXQX at a point. Label this point PP.
  5. Join PP to RR using a ruler. ΔPQR\Delta PQR is the required right-angled triangle.

Explanation:

Since the triangle is right-angled at QQ, QRQR is treated as the base and QXQX is the perpendicular height. The hypotenuse PRPR must connect the far end of the base (RR) to the height (PP). The compass ensures the length PRPR is exactly 5 cm5\text{ cm}.

Problem 2:

Construct ΔLMN\Delta LMN such that ∠M=90∘\angle M = 90^{\circ}, MN=4 cmMN = 4\text{ cm}, and LN=6 cmLN = 6\text{ cm}.

Solution:

  1. Draw the base MN=4 cmMN = 4\text{ cm}.
  2. At vertex MM, construct a 90∘90^{\circ} angle and draw a ray MYMY upwards.
  3. With NN as the center and a radius of 6 cm6\text{ cm}, draw an arc intersecting ray MYMY at point LL.
  4. Join LNLN. The triangle ΔLMN\Delta LMN is constructed.

Explanation:

In this RHS problem, the side MNMN is one leg and LNLN is the hypotenuse (the side opposite the 90∘90^{\circ} angle at MM). We use the compass from NN to find the point LL on the vertical line MYMY.

Problem 3:

Construct a right-angled triangle ΔABC\Delta ABC, right-angled at BB, given that BC=6 cmBC = 6\text{ cm} and AC=10 cmAC = 10\text{ cm}.

Triangle ABC with BC=6cm and AC=10cm.

Solution:

  1. Draw a line segment BC=6 cmBC = 6\text{ cm}.
  2. At point BB, construct a ray BXBX making an angle of 90∘90^{\circ} with BCBC.
  3. With CC as center and radius 10 cm10\text{ cm} (the hypotenuse), draw an arc intersecting ray BXBX at point AA.
  4. Join ACAC to complete the triangle ΔABC\Delta ABC.

Explanation:

Since the triangle is right-angled at BB, ACAC must be the hypotenuse. We use the RHS criterion where R=90∘R=90^{\circ}, H=10 cmH=10\text{ cm}, and S=6 cmS=6\text{ cm}.

Problem 4:

Construct an isosceles right-angled triangle ΔXYZ\Delta XYZ where m∠Y=90∘m\angle Y = 90^{\circ} and the equal sides are 5 cm5\text{ cm} each.

Isosceles right-angled triangle XYZ with legs of 5cm.

Solution:

  1. Draw a line segment YZ=5 cmYZ = 5\text{ cm}.
  2. At point YY, construct a ray YPYP perpendicular to YZYZ (90∘90^{\circ}).
  3. From YY, mark a point XX on ray YPYP such that YX=5 cmYX = 5\text{ cm}.
  4. Join XZXZ.

Explanation:

In an isosceles right triangle, the two legs forming the right angle are equal. Here YX=YZ=5 cmYX = YZ = 5\text{ cm}. The hypotenuse XZXZ is determined by connecting the endpoints of these legs.